Question

先化简 再求值:已知x的平方-2=0,求代数式(x-1)的平方除以(x的平方-1)+x的平方/(x+1)的值

Answer 1

(x-1)²/(x²-1)+x²/(x+1)

=[(x-1)²+x²(x-1)]/(x²-1)

=(x²-2x+1+x³-x²)/(x²-1)

=(-2x+1+x³)/(x²-1)

=[1+x(x²-2)]/(x²-2+1)

=1/1

=1

[(a²-b²)/(a²-ab)]/[a+(2ab+b²)/a]

=[(a²-b²)/(a²-ab)]/[(a²+2ab+b²)/a]

=[(a²-b²)/(a²-ab)]*[a/(a+b)²]

=(a²-b²)(a+b)²/(a-b)

=(a+b)³