Belajar matematika sejak dini
36.Bentuk sederhana dari33+800−27−2162=....(SIMAK UI 2012 Mat IPA)a.2−2b.8−2c.−2+2d.2+52e.8+52Jawab:bmisalkan,x=33+800−27−2162=(33+202−27−2.92)=33+202−27−182x2=33+202+27−182−2(33+202)(27−182)=60+22−233.27−33.182+27.202−20.18.2=60+22−2891−720+5402−5942=60+22−2171−542=60+22−2171−2.272=60+22−2171−227.272=60+22−2171−2162.9=60+22−2162+9−2162.9=60+22−2(162−9)=60+22−2(92−3)x2=66−162x=66−2.82=64+2−264.2=64−2=8−2
37.Jika12+3+5=a2+b3+c512,makaa+b+c=....(UM UGM 2016 Mat Das)a.0b.1c.2d.3e.4Jawab:e12+3+5=12+3+5×(2+3−5)(2+3−5)=2+3−5(2+3)2−(5)2=2+3−52+3+22.3−5=2+3−526×66=12+18−3012=23+32−3012=32+23−3012{a=3b=2c=−1a+b+c=3+2+(−1)=4
38.Tunjukkan bahwa1+21+31+41+51+⋯=3Buktix2=x2x2=1+(x2−1)=1+(x−1)(x+1)=1+(x−1)(x+1)2=1+(x−1)1+(x+1)2−1=1+(x−1)1+(x+1−1)(x+1+1)=1+(x−1)1+x(x+2)=1+(x−1)1+x(x+2)2=1+(x−1)1+x1+(x+2)2−1=1+(x−1)1+x1+(x+2−1)(x+2+1)=1+(x−1)1+x1+(x+1)(x+3)=1+(x−1)1+x1+(x+1)(x+3)2x2=1+(x−1)1+x1+(x+1)...x=1+(x−1)1+x1+(x+1)...
39.Jika bentukab−1a−1−b−1dinyatakan dalam pangkat positif=....a.a2a−bb.a2a−1c.b−aabd.a2b−ae.1a−b(\textit{SAT Test Math Level 2})Jawab:dab−1a−1−b−1=ab−1a−1−b−1×bb=aa−1b−1=aa−1b−1×aa=a2b−a
40.Jika terdapat hubungan berikuta.2p=3q=6r,tunjukkan bahwapr+qr−pq=0b.2x=32y=6z,tunjukkan bahwa 2xy−2yz−xz=0c.315a=55b=153c,tunjukkan bahwa 5ab−bc−3ac=0
bukti
Yang akan ditunjukkan adalah no. 40 yang poin c, yaitu:
315a=55b=153c{3=55b15a315a5b=b(ab=cd→a=cdbatauabd=c)315a=153c315a=(3×5)3c315a=(3×315a5b)3c315a=33c+9cbaf(x)=ag(x)f(x)=g(x)15a=3c+9acb15ab=3bc+9ac5ab=bc+3ac5ab−bc−3ac=0◼
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