(i)
2x(x2+2)3x−1=xA+x2+2Bx+x2+2C=
=2x(x2+2)Ax2+2A+Bx2+Cx A+B=0
C=23
2A=−21
A=−41,B=41,C=23
2x(x2+2)3x−1=x−41+x2+241x+x2+223 (ii)
(x−1)(x2+1)2x+1=x−1A+x2+1Bx+x2+1C=
=(x−1)(x2+1)Ax2+A+Bx2−Bx+Cx−C A+B=0
−B+C=2
A−C=1
A=23,B=−23,C=21
(x−1)(x2+1)2x+1=x−123+x2+1−23x+x2+121
(iii)
x(x2−x+1)x3+x+1=1+x(x2−x+1)x2+1=
=1+xA+x2−x+1Bx+x2−x+1C=
=1+x(x2−x+1)Ax2−Ax+A+Bx2+Cx A+B=1
−A+C=0
A=1
A=1,B=0,C=1
x(x2−x+1)x3+x+1=1+x1+x2−x+11
(iv)
(x−2)(x2+3x+1)3x+2
Let x2+3x+1=0
D=(3)2−4(1)(1)=5
x=2−3±5
x2+3x+1=(x−2−3−5)(x−2−3+5)=
=41(2x+3+5)(2x+3−5)
(x−2)(x2+3x+1)3x+2=x−2A+2x+3+52B+2x+3−52C=
=4(x−2)(x2+3x+1)4Ax2+12Ax+4A+(x−2)(4Bx+6B−25B+4Cx+6C+25C)
4A+4B+4C=0
12A−8B+6B−25B−8C+6C+25C=3
4A−12B+45B−12C−45C=2
A+B+C=0
−12B−12C−2B−25B−2C+25C=3
−16B+45B−16C−45C=2
A+B+C=0
−14B−14C−25B+25C=3
−8B+25B−8C−25C=1
A=112
B=−111−445
C=−111+445
(x−2)(x2+3x+1)3x+2=x−2112+2x+3+5−112−225+2x+3−5−112+225
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