已知(3x-4)/(x^2-3x+2)=a/(x-2)+b/(x-1)是恒等式,求a+b的值
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已知(3x-4)/(x^2-3x+2)=a/(x-2)+b/(x-1)是恒等式,求a+b的值
已知(3x-4)/(x^2-3x+2)=a/(x-2)+b/(x-1)是恒等式,求a+b的值
已知(3x-4)/(x^2-3x+2)=a/(x-2)+b/(x-1)是恒等式,求a+b的值
【数学之美团为你解答】
a/(x-2)+b/(x-1)
=[a(x-1)+b(x-2)]/[(x-2)(x-1)]
=[(a+b)x-(a+2b)]/(x²-3x+2)=(3x-4)/(x²-3x+2)
a+b=3
a+2b=4
b=1,a=2
a/(x-2)+b/(x-1)
=[a(x-1)+b(x-2)]/[(x-2)(x-1)]
=[(a+b)x-(a+2b)]/[x²-3x+2]
得:
a+b=3、a+2b=4