Contoh Soal 8 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{ll}\\ 36.&\textrm{Bentuk sederhana dari}\\ &\sqrt{33+\sqrt{800}}-\sqrt{27-2\sqrt{162}}=....\\\\ &\qquad\qquad\qquad(\textbf{SIMAK UI 2012 Mat IPA})\\ &\begin{array}{llllllll}\\ \textrm{a}.&2-\sqrt{2}\\ \color{red}\textrm{b}.&8-\sqrt{2}\\ \textrm{c}.&-2+\sqrt{2}\\ \textrm{d}.&2+5\sqrt{2}\\ \textrm{e}.&8+5\sqrt{2} \end{array}\\\\ &\textrm{Jawab}:\qquad \color{red}\textbf{b}\\ &\textrm{misalkan},\\ &\color{blue}\begin{aligned}x&=\sqrt{33+\sqrt{800}}-\sqrt{27-2\sqrt{162}}\\ &=\left ( \sqrt{33+20\sqrt{2}}-\sqrt{27-2.9\sqrt{2}}\: \right )\\ &=\sqrt{33+20\sqrt{2}}-\sqrt{27-18\sqrt{2}}\\ x^{2}&=33+20\sqrt{2}+27-18\sqrt{2}-2\sqrt{\left ( 33+20\sqrt{2} \right )\left ( 27-18\sqrt{2} \right )}\\ &=60+2\sqrt{2}-2\sqrt{33.27-33.18\sqrt{2}+27.20\sqrt{2}-20.18.2}\\ &=60+2\sqrt{2}-2\sqrt{891-720+540\sqrt{2}-594\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-54\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-2.27\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-2\sqrt{27.27}\sqrt{2}}\\ &=60+2\sqrt{2}-2\sqrt{171-2\sqrt{162.9}}\\ &=60+2\sqrt{2}-2\sqrt{162+9-2\sqrt{162.9}}\\ &=60+2\sqrt{2}-2\left ( \sqrt{162}-\sqrt{9} \right )\\ &=60+2\sqrt{2}-2\left ( 9\sqrt{2}-3 \right )\\ x^{2}&=66-16\sqrt{2}\\ x&=\sqrt{66-2.8\sqrt{2}}\\ &=\sqrt{64+2-2\sqrt{64.2}}\\ &=\sqrt{64}-\sqrt{2}\\ &=8-\sqrt{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 37.&\textrm{Jika}\: \: \displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\displaystyle \frac{a\sqrt{2}+b\sqrt{3}+c\sqrt{5}}{12}\: ,\\\\ &\textrm{maka}\: \: a+b+c=....\\ &\qquad\qquad\qquad\qquad (\textbf{UM UGM 2016 Mat Das})\\ &\begin{array}{llllllll}\\ \textrm{a}.&0\\ \textrm{b}.&1\\ \textrm{c}.&2\\ \textrm{d}.&3\\ \color{red}\textrm{e}.&4 \end{array}\\\\ &\textrm{Jawab}:\qquad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}&=\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\times \displaystyle \frac{\left ( \sqrt{2}+\sqrt{3}-\sqrt{5} \right )}{\left ( \sqrt{2}+\sqrt{3}-\sqrt{5} \right )}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\left ( \sqrt{2}+\sqrt{3} \right )^{2}-\left ( \sqrt{5} \right )^{2}}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2+3+2\sqrt{2.3}-5}\\ &=\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2\sqrt{6}}\times \displaystyle \frac{\sqrt{6}}{\sqrt{6}}\\ &=\displaystyle \frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{12}\\ &=\displaystyle \frac{2\sqrt{3}+3\sqrt{2}-\sqrt{30}}{12}\\ &=\displaystyle \frac{3\sqrt{2}+2\sqrt{3}-\sqrt{30}}{12}\\ &\quad \begin{cases} a &=3 \\ b &=2 \\ c &=-1 \end{cases}\\ a+b+c&=3+2+(-1)\\ &=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 38.&\textrm{Tunjukkan bahwa}\\ &\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+\cdots }}}}}=3\\\\ &\textrm{Bukti}\\ &\color{blue}\begin{aligned}x^{2}&=x^{2}\\ x^{2}&=1+\left ( x^{2}-1 \right )\\ &=1+\left ( x-1 \right )\left ( x+1 \right )\\ &=1+\left ( x-1 \right )\sqrt{\left ( x+1 \right )^{2}}\\ &=1+\left ( x-1 \right )\sqrt{1+\left ( x+1 \right )^{2}-1}\\ &=1+\left ( x-1 \right )\sqrt{1+\left ( x+1-1 \right )\left ( x+1+1 \right )}\\ &=1+\left ( x-1 \right )\sqrt{1+x\left ( x+2 \right )}\\ &=1+\left ( x-1 \right )\sqrt{1+x\sqrt{\left ( x+2 \right )^{2}}}\\ &=1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+2 \right )^{2}-1}}\\ &=1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+2-1 \right )\left ( x+2+1 \right )}}\\ &=1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+1 \right )\left ( x+3 \right )}}\\ &=1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+1 \right )\sqrt{\left ( x+3 \right )^{2}}}}\\ x^{2}&=1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+1 \right )\sqrt{...}}}\\ x&=\sqrt{1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+1 \right )\sqrt{...}}}} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 39.&\textrm{Jika bentuk}\: \: \displaystyle \frac{ab^{-1}}{a^{-1}-b^{-1}}\\ & \textrm{dinyatakan dalam pangkat positif}=\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle \frac{a^{2}}{a-b}&&&\\\\ \textrm{b}.&\displaystyle \frac{a^{2}}{a-1}\\\\ \textrm{c}.&\displaystyle \frac{b-a}{ab}\\\\ \color{red}\textrm{d}.&\displaystyle \frac{a^{2}}{b-a}\\\\ \textrm{e}.&\displaystyle \frac{1}{a-b}\\\\ &&&&(\textbf{\textit{SAT Test Math Level 2}})\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\displaystyle \frac{ab^{-1}}{a^{-1}-b^{-1}}&=\displaystyle \frac{ab^{-1}}{a^{-1}-b^{-1}}\times \frac{b}{b}\\ &=\displaystyle \frac{a}{a^{-1}b-1}\\ &=\displaystyle \frac{a}{a^{-1}b-1}\times \frac{a}{a}\\ &=\displaystyle \frac{a^{2}}{b-a} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 40.&\textrm{Jika terdapat hubungan berikut}\\ &\textrm{a}.\quad 2^{p}=3^{q}=6^{r},\: \: \textrm{tunjukkan bahwa}\: \: pr+qr-pq=0\\ &\textrm{b}.\quad 2^{x}=3^{2y}=6^{z},\: \: \textrm{tunjukkan bahwa }\: \: 2xy-2yz-xz=0\\ &\textrm{c}.\quad 3^{15a}=5^{5b}=15^{3c},\: \: \textrm{tunjukkan bahwa }\: \: 5ab-bc-3ac=0\\ \end{array}$

$\textbf{bukti}$

Yang akan ditunjukkan adalah no. 40 yang poin c, yaitu:

$\color{blue}\begin{aligned}3^{15a}&=5^{5b}=15^{3c}\begin{cases} 3=5^{\frac{5b}{15a}} & \\ 3^{\frac{15a}{5b}}=b &\left ( a^{b}=c^{d}\rightarrow a=c^{\frac{d}{b}}\: \: \textrm{atau}\: \: a^{\frac{b}{d}}=c \right ) \end{cases}\\ 3^{15a}&=15^{3c}\\ 3^{15a}&=(3\times 5)^{3c}\\ 3^{15a}&=(3\times 3^{\frac{15a}{5b}})^{3c}\\ 3^{15a}&=3^{3c+\frac{9c}{b}}\\ a^{f(x)}&=a^{g(x)}\\ f(x)&=g(x)\\ 15a&=3c+\frac{9ac}{b}\\ 15ab&=3bc+9ac\\ 5ab&=bc+3ac\\ 5ab-bc-3ac&=0\quad \color{black}\blacksquare \end{aligned}$