Lanjutan 3 (Barisan & Deret)

$\begin{array}{l}\\ 11.& \text{Hasil penjumlahan dari}\\ & {\displaystyle \frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\cdots +\frac{1}{97.99}=\cdots }\\ & \begin{array}{l}\\ \text{A}.& {\displaystyle \frac{98}{99}}& & & \text{D}.& {\displaystyle \frac{48}{99}}\\ \\ \text{B}.& {\displaystyle \frac{50}{99}}& \text{C}.& {\displaystyle \frac{49}{99}}& \text{E}.& {\displaystyle \frac{47}{99}}\end{array}\\ \\ & \mathbf{\text{Pembahasan}}:\\ & \begin{array}{rl}& \text{Bentuk di atas memenuhi bentuk}\\ & {\displaystyle \frac{1}{x(x+2)}={\displaystyle \frac{1}{2}\left({\displaystyle \frac{1}{x}-\frac{1}{(x+2)}}\right).\text{Bentuk ini pada}}}\\ & \text{bilangan dengan pola tertentu seperti di atas akan}\\ & \text{menghabiskan dengan bilangan sebelahnya}\\ & \text{atau lazim dikenal dengan}\mathbf{\text{prinsip teleskoping}}\\ & \text{Sebagaimana bentuk penjumlahan dengan pola di atas}\\ & \text{maka}\\ & ={\displaystyle \frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\cdots +\frac{1}{97}-\frac{1}{99})}\\ & ={\displaystyle \frac{1}{2}(1-\frac{1}{99})={\displaystyle \frac{1}{2}.\frac{98}{99}={\displaystyle \frac{49}{99}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 12.& \text{Diketahui}\\ & x={\displaystyle \frac{1+p+p^2+p^3+\cdots +p^{n-1}}{1+p+p^2+p^3+\cdots +p^{n-2}+p^{n-1}+p^n}}\\ & y={\displaystyle \frac{1+q+q^2+q^3+\cdots +q^{n-1}}{1+q+q^2+q^3+\cdots +q^{n-2}+q^{n-1}+q^n}}\\ \\ & \text{dan}p>q>0\\ \\ & \text{Tunjukkan bahwa}x<y\\ \\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}& \text{Perhatikan bahwa}:p>q>0\\ & \text{sehingga}\\ & {\displaystyle \frac{1}{p}<\frac{1}{q},{\displaystyle \frac{1}{p^2}<\frac{1}{q^2},\cdots ,{\displaystyle \frac{1}{p^n}<\frac{1}{q^n}}}}\\ & \text{Jika bentuk di atas dijumlahkan, maka}\\ & {\displaystyle \frac{1}{p}+\frac{1}{p^2}+\cdots +\frac{1}{p^n}<\frac{1}{q}+\frac{1}{q^2}+\cdots +\frac{1}{q^n}}\\ & \iff {\displaystyle \frac{p^{n-1}+\cdots +p^2+p+1}{p^n}<{\displaystyle \frac{q^{n-1}+\cdots +q^2+q+1}{q^n}}}\\ & \iff {\displaystyle \frac{p^n}{1+p+p^2+\cdots +p^{n-1}}>{\displaystyle \frac{q^n}{1+q+q^2+\cdots +q^{n-1}}}}\\ & \iff {\displaystyle \frac{p^n}{1+p+p^2+\cdots +p^{n-1}}+1>{\displaystyle \frac{q^n}{1+q+q^2+\cdots +q^{n-1}}+1}}\\ & \iff {\displaystyle \frac{1+p+p^2+\cdots +p^{n-1}+p^n}{1+p+p^2+\cdots +p^{n-1}}>{\displaystyle \frac{1+q+q^2+\cdots +q^{n-1}+q^n}{1+q+q^2+\cdots +q^{n-1}}}}\\ & \iff {\displaystyle \frac{1+p+p^2+\cdots +p^{n-1}}{1+p+p^2+\cdots +p^{n-1}+p^n}<{\displaystyle \frac{1+q+q^2+\cdots +q^{n-1}}{1+q+q^2+\cdots +q^{n-1}+q^n}}}\\ & \iff x<y◼\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Aziz, A. 2016. Rahasia Juara Olimpiade Matematika SMA. Yogyakarta: ANDI.
2. Baskoro, B.D. 2012. Aljabar dan Trigonometri Cespleng Olimpiade Matematika. Yogyakarta: BERLIAN.
3. Bintari, N., Gunarto, D. 2007. Panduan Menguasai Soal-Soal Olimpiade Nasional & Internasional. Yogyakarta: INDONESIA CERDAS.
4. Idris, M,. Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
5. Sembiring, S. 2002. Olimpiade Matematika Untuk SMU. Bandung: YRAMA WIDYA.
6. Sembiring, S., Suparmin, S. 2015. Pena Emas OSN Matematika SMA. Bandung: YRAMA WIDYA.

Lanjutan 2 (Barisan & Deret)

$\begin{array}{l}\\ 9.& \text{Tentukan hasil dari}\\ & {\displaystyle \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\frac{9}{32}+\cdots }\\ \\ & \mathbf{\text{Pembahasan}}:\\ & \mathbf{\text{Alternatif 1}}\\ & \begin{array}{rl}& \text{Misalkan}S={\displaystyle \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\frac{9}{32}+\cdots }\\ & \text{maka}{\displaystyle \frac{1}{2}S={\displaystyle \frac{1}{4}+\frac{3}{8}+\frac{5}{16}+\frac{7}{32}+\frac{9}{64}+\cdots }}\\ & \text{Jika}\\ & S-{\displaystyle \frac{1}{2}S={\displaystyle \frac{1}{2}+\frac{2}{4}+\frac{2}{8}+\frac{2}{16}+\frac{2}{32}+\frac{2}{64}+\cdots }}\\ & \begin{array}{rl}\iff {\displaystyle \frac{1}{2}S}& ={\displaystyle \frac{1}{2}+\underset{\text{deret geometri}}{\underbrace{(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\cdots )}}}\\ \iff {\displaystyle \frac{1}{2}S}& ={\displaystyle \frac{1}{2}+{\displaystyle \frac{{\displaystyle \frac{1}{2}}}{1-{\displaystyle \frac{1}{2}}}={\displaystyle \frac{1}{2}+{\displaystyle \frac{\frac{1}{2}}{\frac{1}{2}}={\displaystyle \frac{1}{2}+1={\displaystyle \frac{3}{2}}}}}}}\\ \iff {\displaystyle \frac{1}{2}S}& ={\displaystyle \frac{3}{2}}\\ \iff S& =3\end{array}\end{array}\\ & \mathbf{\text{Alternatif 2}}\\ & \begin{array}{rl}& S_{\mathrm{\infty }}={\displaystyle \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\frac{9}{32}+\cdots }\\ & \text{dengan suku awal geometri bukan 1, maka kita ubah menjadi}\\ & 2S_{\mathrm{\infty }}=1+{\displaystyle \frac{3}{2}+\frac{5}{4}+\frac{7}{8}+\frac{9}{16}+\cdots }\\ & \begin{array}{|cc|}\hline \text{Bagian aritmetika}& \text{Bagian Geometri}\\ (\text{lihat bagian pembilang})& (\text{bagian pembilang-penyebut})\\ \{\begin{array}{ll}\bullet a& =U_1=1\\ \bullet b& =U_2-U_1=3-1=2\end{array}& \bullet r={\displaystyle \frac{U_2}{U_1}=\frac{\frac{1}{2}}{1}={\displaystyle \frac{1}{2}}}\\ \hline\end{array}\\ & \begin{array}{rl}2S_{\mathrm{\infty }}& ={\displaystyle \frac{a}{(1-r)}+\frac{br}{(1-r{)}^2}}\\ 2S_{\mathrm{\infty }}& ={\displaystyle \frac{1}{1-\frac{1}{2}}+\frac{2\times \frac{1}{2}}{(1-\frac{1}{2}{)}^2}=2+4=6}\\ S_{\mathrm{\infty }}& =3\end{array}\\ & \text{Jadi},{\displaystyle \frac{1}{2}+\frac{3}{4}+\frac{5}{8}+\frac{7}{16}+\frac{9}{32}+\cdots =3}\end{array}\end{array}$.

$.\text{Untuk link materi pada pembahasan alternatif 2}$. di sini

$\begin{array}{l}\\ 10.& \text{Hasil penjumlahan dari}\\ & {\displaystyle \frac{1}{3}+\frac{2}{9}+\frac{3}{27}+\frac{4}{81}+\frac{5}{243}+\cdots =\cdots }\\ & \begin{array}{l}\\ \text{A}.& {\displaystyle \frac{2}{3}}& & & \text{D}.& {\displaystyle \frac{4}{3}}\\ \\ \text{B}.& {\displaystyle \frac{3}{4}}& \text{C}.& 1& \text{E}.& {\displaystyle \frac{3}{2}}\end{array}\\ \\ & \mathbf{\text{Pembahasan}}:\\ & \begin{array}{rl}& S_{\mathrm{\infty }}={\displaystyle \frac{1}{3}+\frac{2}{9}+\frac{3}{27}+\frac{4}{81}+\frac{5}{243}+\cdots }\\ & \text{dengan suku awal geometri bukan 1, maka kita ubah menjadi}\\ & 3S_{\mathrm{\infty }}=1+{\displaystyle \frac{2}{3}+\frac{3}{9}+\frac{4}{27}+\frac{5}{81}+\cdots }\\ & \begin{array}{|cc|}\hline \text{Bagian aritmetika}& \text{Bagian Geometri}\\ (\text{lihat bagian pembilang})& (\text{bagian pembilang-penyebut})\\ \{\begin{array}{ll}\bullet a& =U_1=1\\ \bullet b& =U_2-U_1=2-1=1\end{array}& \bullet r={\displaystyle \frac{U_2}{U_1}=\frac{\frac{1}{3}}{1}={\displaystyle \frac{1}{3}}}\\ \hline\end{array}\\ & \begin{array}{rl}3S_{\mathrm{\infty }}& ={\displaystyle \frac{a}{(1-r)}+\frac{br}{(1-r{)}^2}}\\ 3S_{\mathrm{\infty }}& ={\displaystyle \frac{1}{1-\frac{1}{3}}+\frac{1\times \frac{1}{3}}{(1-\frac{1}{3}{)}^2}={\displaystyle \frac{3}{2}+\frac{3}{4}={\displaystyle \frac{9}{4}}}}\\ S_{\mathrm{\infty }}& ={\displaystyle \frac{3}{4}}\end{array}\\ & \text{Jadi},{\displaystyle \frac{1}{3}+\frac{2}{9}+\frac{3}{27}+\frac{4}{81}+\frac{5}{243}+\cdots ={\displaystyle \frac{3}{4}}}\end{array}\end{array}$ .

Lanjutan (Barisan & Deret)

$\begin{array}{l}\\ 5.& \mathbf{\text{(Lomba Matematika Nasional}}\\ & \mathbf{\text{HIMATIKA UGM 2006)}}\\ & \text{Jika bilangan}\\ & A={\displaystyle \frac{1}{1+1}+\frac{1}{1+2}+\frac{1}{1+3}+\cdots +\frac{1}{1+100}}\\ & B={\displaystyle \frac{1}{1+1}+\frac{1}{1+{\displaystyle \frac{1}{2}}}+\frac{1}{1+{\displaystyle \frac{1}{3}}}+\cdots +\frac{1}{1+{\displaystyle \frac{1}{100}}}}\\ & \text{maka}A+B\text{sama dengan}....\\ & \text{A}.202\text{D}.100\\ & \text{B}.200\text{C}.101\text{E}.99\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan bahwa}\\ & \begin{array}{l}\\ \begin{array}{rl}A& ={\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots +\frac{1}{101}}\\ B& ={\displaystyle \frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\cdots +\frac{100}{101}}\end{array}& \\ & +\\ \\ A+B=\underset{100}{\underbrace{1+1+1+\cdots +1}}& =100\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \mathbf{\text{(OSN tk. Kota/Kab 2002)}}\\ & \text{Misalkan}\\ & a={\displaystyle \frac{1^2}{1}+\frac{2^2}{3}+\frac{3^2}{5}+\cdots +\frac{{1001}^2}{2001}}\\ \\ & b={\displaystyle \frac{1^2}{3}+\frac{2^2}{5}+\frac{3^2}{7}+\cdots +\frac{{1001}^2}{2003}}\\ & \text{Tentukanlah bilangan bulat yang}\\ & \text{nilainya paling dekat ke}a-b\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan bahwa}\\ & \begin{array}{rl}a-b& =\left({\displaystyle \frac{1^2}{1}+\frac{2^2}{3}+\frac{3^2}{5}+\cdots +\frac{{1001}^2}{2001}}\right)\\ & -\left({\displaystyle \frac{1^2}{3}+\frac{2^2}{5}+\frac{3^2}{7}+\cdots +\frac{{1001}^2}{2003}}\right)\\ & ={\displaystyle \frac{1^2}{1}+\left({\displaystyle \frac{2^2}{3}-{\displaystyle \frac{1^2}{3}}}\right)+\left({\displaystyle \frac{3^2}{5}-{\displaystyle \frac{2^2}{5}}}\right)}\\ & +\left({\displaystyle \frac{4^2}{7}-{\displaystyle \frac{3^2}{7}}}\right)+\cdots +\left({\displaystyle \frac{{1001}^2}{2001}-{\displaystyle \frac{{1000}^2}{2001}}}\right)\\ & -{\displaystyle \frac{{1001}^2}{2003}}\\ & =\underset{1001}{\underbrace{1+1+1+\cdots +1}}-{\displaystyle \frac{{1001}^2}{2003}}\\ & =1001-{\displaystyle \frac{{1001}^2}{2003}=1001\left({\displaystyle \frac{2003-1001}{2003}}\right)}\\ & ={\displaystyle \frac{1001\times 1002}{2003}>{\displaystyle \frac{1001}{2}=500,5}}\end{array}\\ & \text{Jadi bilangan bulat yang paling dekat}\\ & \text{ke}a-b\text{adalah}501\end{array}$.

 $\begin{array}{l}\\ 7.& \text{Misalkan}\\ & a_n={\displaystyle \frac{n(n+1)}{2},\text{tentukanlah jumlah}}\\ & \text{dari}{\displaystyle \frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_{2023}}}\\ \\ & \mathbf{\text{Pembahasan}}:\\ & \begin{array}{rl}{a}_n& ={\displaystyle \frac{n(n+1)}{2},\text{maka}}\\ {\displaystyle \frac{1}{a_n}}& ={\displaystyle \frac{2}{n(n+1)}}\\ & =2\left({\displaystyle \frac{1}{n}-\frac{1}{n+1}}\right)\\ & \text{lihat pembahasan no.3 di atas}\\ & \end{array}\\ & \begin{array}{rl}& {\displaystyle \frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_{2023}}}\\ & =2((1-{\displaystyle \frac{1}{2}})+\left({\displaystyle \frac{1}{2}-\frac{1}{3}}\right)+\cdots +\left({\displaystyle \frac{1}{2022}-\frac{1}{2023}}\right))\\ & =2(1-{\displaystyle \frac{1}{2023}})\\ & =2\left({\displaystyle \frac{2022}{2023}}\right)={\displaystyle \frac{4044}{2023}}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Misalkan}n\text{adalah bilangan asli}\\ & \text{dan}\left\{a_n\right\}\text{adalah barisan bilangan real}\\ & \text{dengan}{a}_n={\displaystyle \frac{2^n}{2^{2n+1}-2^{n+1}-2^n+1}}\\ \\ & \text{Tunjukkan bahwa untuk setiap bilangan}\\ & \text{asli}n,\text{berlaku}{a}_1+a_2+\cdots +a_n<1\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}{a}_n& ={\displaystyle \frac{2^n}{2^{2n+1}-2^{n+1}-2^n+1}}\\ & ={\displaystyle \frac{2^n}{(2^{n+1}-1)(2^n-1)}}\\ & =\left({\displaystyle \frac{1}{2^{n+1}-1}-\frac{1}{2^n-1}}\right)\\ a_1& ={\displaystyle \frac{1}{2^1-1}-\frac{1}{2^2-1}=1-{\displaystyle \frac{1}{3}}}\\ a_2& ={\displaystyle \frac{1}{2^2-1}-\frac{1}{2^3-1}=\frac{1}{3}-\frac{1}{7}}\\ a_3& ={\displaystyle \frac{1}{2^3-1}-\frac{1}{2^4-1}=\frac{1}{7}-\frac{1}{15}}\\ & ⋮⋮\\ a_n& ={\displaystyle \frac{1}{2^n-1}-\frac{1}{2^{n+1}-1}+}\\ a_1& +a_2+a_3+\cdots +a_n\\ & =1-{\displaystyle \frac{1}{2^{n+1}-1}<1◼}\end{array}\end{array}$.

Selingan (Barisan & Deret)

Lanjutan contoh soal dan pembahasannya terkait barisan dan deret

$\begin{array}{l}\\ 1.& \text{Hasil kali bilangan bentuk berikut}\\ & (1-{\displaystyle \frac{1}{4}})(1-{\displaystyle \frac{1}{5}})(1-{\displaystyle \frac{1}{6}})\cdots (1-{\displaystyle \frac{1}{100}})\\ & \text{adalah}....\\ & \text{A}.{\displaystyle \frac{1}{100}\text{D}.{\displaystyle \frac{4}{100}}}\\ \\ & \text{B}.{\displaystyle \frac{2}{100}\text{C}.{\displaystyle \frac{3}{100}\text{E}.{\displaystyle \frac{5}{100}}}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan bahwa}\\ & \begin{array}{rl}& (1-{\displaystyle \frac{1}{4}})(1-{\displaystyle \frac{1}{5}})(1-{\displaystyle \frac{1}{6}})\cdots (1-{\displaystyle \frac{1}{100}})\\ & =\left({\displaystyle \frac{3}{4}}\right)\left({\displaystyle \frac{4}{5}}\right)\left({\displaystyle \frac{5}{6}}\right)\cdots \left({\displaystyle \frac{99}{100}}\right)\\ & =\left({\displaystyle \frac{3}{\text{⧸}4}}\right)\left({\displaystyle \frac{\text{⧸}4}{\text{⧸}5}}\right)\left({\displaystyle \frac{\text{⧸}5}{\text{⧸}6}}\right)\cdots \left({\displaystyle \frac{\text{⧸}99}{100}}\right)\\ & ={\displaystyle \frac{3}{100}}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Hasil kali bilangan bentuk berikut}\\ & (1-{\displaystyle \frac{2}{3}})(1-{\displaystyle \frac{2}{5}})(1-{\displaystyle \frac{2}{7}})\cdots (1-{\displaystyle \frac{2}{2023}})\\ & \text{adalah}....\\ & \text{A}.{\displaystyle \frac{1}{2023}\text{D}.{\displaystyle \frac{4}{2023}}}\\ \\ & \text{B}.{\displaystyle \frac{2}{2023}\text{C}.{\displaystyle \frac{3}{2023}\text{E}.{\displaystyle \frac{5}{2023}}}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Dengan cara pembahasan pada no.1 di atas, maka}\\ & \begin{array}{rl}& (1-{\displaystyle \frac{2}{3}})(1-{\displaystyle \frac{2}{5}})(1-{\displaystyle \frac{2}{7}})\cdots (1-{\displaystyle \frac{2}{2023}})\\ & =\left({\displaystyle \frac{1}{3}}\right)\left({\displaystyle \frac{3}{5}}\right)\left({\displaystyle \frac{5}{7}}\right)\cdots \left({\displaystyle \frac{2021}{2023}}\right)\\ & =\left({\displaystyle \frac{1}{\text{⧸}3}}\right)\left({\displaystyle \frac{\text{⧸}3}{\text{⧸}5}}\right)\left({\displaystyle \frac{\text{⧸}5}{\text{⧸}7}}\right)\cdots \left({\displaystyle \frac{\text{⧸}2021}{2023}}\right)\\ & ={\displaystyle \frac{1}{2023}}\end{array}\end{array}$.

 $\begin{array}{l}\\ 3.& \text{Bentuk sederhana dari}\\ & {\displaystyle \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{2022\times 2023}}\\ \\ & \mathbf{\text{Pembahasan}}:\\ & \begin{array}{rl}& =(1-{\displaystyle \frac{1}{2}})+\left({\displaystyle \frac{1}{2}-\frac{1}{3}}\right)+\left({\displaystyle \frac{1}{3}-\frac{1}{4}}\right)+\\ & +\cdots +\left({\displaystyle \frac{1}{2022}-\frac{1}{2023}}\right)\\ & =1-{\displaystyle \frac{1}{2023}}\\ & ={\displaystyle \frac{2022}{2023}}\end{array}\end{array}$.

$\begin{array}{|c|}\hline \begin{array}{rl}\mathbf{\text{Seb}}& \mathbf{\text{agai pengingat kita}}\\ \bullet & 1+2+3+\cdots +n={\displaystyle \frac{n(n+1)}{2}}\\ \bullet & 1+3+5+\cdots +(2n-1)=n^2\\ \bullet & 1^2+2^2+3^2+\cdots +n^2={\displaystyle \frac{n(n+1)(2n+1)}{6}}\\ \bullet & 1^3+2^3+3^3+\cdots +n^3={\left({\displaystyle \frac{n(n+1)}{2}}\right)}^2\\ \bullet & 1+{\displaystyle \frac{1}{1+2}+\frac{1}{1+2+3}+\cdots +\frac{1}{1+2+3+\cdots +n}={\displaystyle \frac{2n}{n+1}}}\\ \bullet & {\displaystyle \frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots +\frac{1}{n(n+1)}={\displaystyle \frac{n}{n+1}}}\\ \bullet & {\displaystyle \frac{1}{1.2.3}+\frac{1}{2.3.4}+\cdots +\frac{1}{n(n+1)(n+2)}={\displaystyle \frac{n(n+3)}{4(n+1)(n+2)}}}\end{array}\\ \hline\end{array}$.

$\begin{array}{l}\\ 4.& \text{Bentuk sederhana dari}\\ & {\displaystyle \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots +\frac{1}{\sqrt{2023}-\sqrt{2022}}}\\ \\ & \mathbf{\text{Pembahasan}}:\\ & \begin{array}{rl}{\displaystyle \frac{1}{1+\sqrt{2}}}& =\frac{1}{1+\sqrt{2}}\times \frac{1-\sqrt{2}}{1-\sqrt{2}}\\ & ={\displaystyle \frac{1-\sqrt{2}}{1^2-(\sqrt{2}{)}^2}=\frac{1-\sqrt{2}}{1-2}}\\ & ={\displaystyle \frac{1-\sqrt{2}}{-1}=\sqrt{2}-1}\\ {\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}}}& =\frac{1}{\sqrt{2}+\sqrt{3}}\times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}\\ & ={\displaystyle \frac{\sqrt{2}-\sqrt{3}}{(\sqrt{2}{)}^2-(\sqrt{3}{)}^2}=\frac{\sqrt{2}-\sqrt{3}}{2-3}}\\ & ={\displaystyle \frac{\sqrt{2}-\sqrt{3}}{-1}=\sqrt{3}-\sqrt{2}}\\ {\displaystyle \frac{1}{\sqrt{3}+\sqrt{4}}}& =\frac{1}{\sqrt{3}+\sqrt{4}}\times \frac{\sqrt{3}-\sqrt{4}}{\sqrt{3}-\sqrt{4}}\\ & ={\displaystyle \frac{\sqrt{3}-\sqrt{4}}{(\sqrt{3}{)}^2-(\sqrt{4}{)}^2}=\frac{\sqrt{3}-\sqrt{4}}{3-4}}\\ & ={\displaystyle \frac{\sqrt{3}-\sqrt{4}}{-1}=\sqrt{4}-\sqrt{3}}\\ & ⋮\end{array}\\ & \text{Sehingga}\\ & \begin{array}{rl}& =\text{⧸}\sqrt{2}-1+\text{⧸}\sqrt{3}-\text{⧸}\sqrt{2}+\text{⧸}\sqrt{4}-\text{⧸}\sqrt{3}+\cdots +\sqrt{2023}-\text{⧸}\sqrt{2022}\\ & =\sqrt{2023}-1\end{array}\end{array}$

Contoh Soal Persiapan Menghadapi Semester Gasal Tahun 2022 untuk Kelas XII MA/SMA/MAK Matematika Wajib (K13 Revisi)

  MATEMATIKA WAJIB

Kelas XII

1. Bangun ruang / dimensi tiga

2. Statistika

Contoh soal dan pembahasan materi terkait di atas lebih dari 85 soal disertai dengan pembahasannya, berikut linknya 

Dimensi tiga

no. 1 sampai dengan 10

no. 11 sampai dengan 20

no. 21 sampai dengan 30

Statistika

no. 1 sampai dengan 5

no. 6 sampai dengan 10

no. 11 sampai dengan 15

no. 16 sampai dengan 20

no. 21 sampai dengan 25

no. 26 sampai dengan 30

no. 31 sampai dengan 35

no. 36

no. 37 sampai dengan 40

no. 41 sampai dengan 45

no. 46 sampai dengan 50

no. 51 sampai dengan 55

no. 56 sampai dengan 59

Contoh Soal Persiapan Menghadapi Semester Gasal Tahun 2022 untuk Kelas XI MA/SMA/MAK Matematika Wajib (K13 Revisi)

 MATEMATIKA WAJIB

Kelas XI

1. Induksi matematika dan pola bilangan

2. Program linear

3. Matrik dan operasi matriks

4. Transformasi geometri

Contoh soal dan pembahasan materi terkait di atas ada sebanyak 100 contoh soal diserati dengan pembahasannya, berikut link soal yang dimaksud

no. 1 sampai dengan 10

no. 11 sampai dengan 20

no. 21 sampai dengan 30

no. 31 sampai dengan 40

no. 41 sampai dengan 50

no. 51 sampai dengan 60

no. 61 sampai dengan 70

no. 71 sampai dengan 80

no. 81 sampai dengan 90

no. 91 sampai dengan 100

Tambahan contoh soal khusus pola bilangan dan barisan serta deret aritmetika dan geometri

no. 1 sampai dengan 5

no. 6 sampai dengan 10

no. 11 sampai dengan 15

no. 16 sampai dengan 20

no. 21 sampai dengan 25

no. 26 sampai dengan 30

no. 31 sampai dengan 35

Tambahan contoh soal khusus transformasi geometri

no. 1 sampai dengan 5

no. 6 sampai dengan 10

no. 11 sampai dengan 15

no. 16 sampai dengan 20

no. 20 sampai dengan 22

Contoh Soal Persiapan Menghadapi Semester Gasal Tahun 2022 untuk Kelas X MA/SMA/MAK Matematika Wajib (K13 Revisi)

MATEMATIKA WAJIB

Kelas X

1. Persamaan dan pertidaksamaan nilai mutlak

2. Pertidaksamaan rasional dan irasional

3. Sistem persamaan linear tiga variabel dan sistem pertidaksamaan dua variabel

Contoh soal dan pembahasan materi terkait di atas ada sebanyak 110 soal disertai dengan pembahasannya, berikut link soalnya

no. 1 sampai dengan 10

no. 11 sampai dengan 15

no. 16 sampai dengan 25

no. 26 sampai dengan 35

no. 36 sampai dengan 45

no. 46 sampai dengan 55

no. 56 sampai dengan 65

no. 66 sampai dengan 75

no. 76 sampai dengan 85

no. 86 sampai dengan 95

no. 96 sampai dengan 105

no. 106 sampai dengan 110