Question

先化简,再求值:[1/（x-3）]×[（x的三次方-6x²+9x）/（x²-2x)]-[(1-x)/(2-x)],其中x=-6

Answer 1

[1/（x-3）]×[（x³-6x²+9x）/（x²-2x)]-[(1-x)/(2-x)]
=1/(x-3)* x*(x-3)²/[x(x-2)] - (1-x)/(2-x)
=(x-3)/(x-2)-(1-x)/(2-x)
=-2/(x-2)
=-2/(-6-2)
=1/4

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Answer 2

[1/（x-3）]×[（x的三次方-6x²+9x）/（x²-2x)]-[(1-x)/(2-x)]
=[1/(x-3)] × [x(x-3)^2/x(x-2)] - [(1-x)/(2-x)]
=[(x-3)/(x-2)] - [(x-1)/(x-2)]
=(x-3-x+1)/(x-2)
= -2/(x-2)
= -2/(-6-2)
= 1/4

Answer 3

[1/（x-3）]×[（x的三次方-6x²+9x）/（x²-2x)]-[(1-x)/(2-x)]=[1/(x-3)] × [x(x-3)^2/x(x-2)] - [(1-x)/(2-x)]=[(x-3)/(x-2)] - [(x-1)/(x-2)]=(x-3-x+1)/(x-2)= -2/(x-2)= -2/(-6-2)= 1/4