已知x2-3x+2=0,那么代数式(x-1)3-x2+1/x-1的值为?代数式意为x-1分之x-1的三次方减去x的平方加1

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已知x2-3x+2=0,那么代数式(x-1)3-x2+1/x-1的值为?代数式意为x-1分之x-1的三次方减去x的平方加1
已知x2-3x+2=0,那么代数式(x-1)3-x2+1/x-1的值为?
代数式意为x-1分之x-1的三次方减去x的平方加1

已知x2-3x+2=0,那么代数式(x-1)3-x2+1/x-1的值为?代数式意为x-1分之x-1的三次方减去x的平方加1
[(x-1)³-x2+1】/(x-1)
=(x-1)²-(x+1)
=x²-3x
由x2-3x+2=0得x2-3x=-2
∴原式=-2

∵(x-1)³-x²+1=(x-1)³-(x²-1)
=(x-1)(x-1)²-(x²-1)
=(x-1)(x²-2x+1)-(x-1)(x+1)
=(x-1)[(x²-2x+1)-(x+1)]
...

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∵(x-1)³-x²+1=(x-1)³-(x²-1)
=(x-1)(x-1)²-(x²-1)
=(x-1)(x²-2x+1)-(x-1)(x+1)
=(x-1)[(x²-2x+1)-(x+1)]
=(x-1)(x²-3x)
∴[(x-1)3-x2+1]/x-1=x²-3x
又x²-3x+2=0,移项有:x²-3x=-2
代数式=-2

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