Question

先化简,再求值:(x+1)(x-1)(x²+x+1)(x²-x-1),其中,x=根号2

实在抱歉………………………………本人有一个地方输错了…………太太对不起那些算出来的人了…………（x²-x-1)应该改为（x²-x+1)!!!!!!

Answer 1

原式=(x^2-1)[x^2+(x+1)][x^2-(x+1)]
=(x^2-1)[x^4-(x+1)^2]
=(x^2-1)(x^4-x^2-2x-1)
因为x=√2 所以x^2=2 x^4=4
所以原式=(2-1)(4-2-2√2-1 )=1-2√2
希望能帮到你啦^0^

Answer 2

解：原式=（（X+1)(X^2+X+1))((X-1)(X^2-X+1)
=(X^3+1)(X^3-1)
=X^6-1
当x=√2时，原式=√2^6-1=8-1=7

Answer 3

原式等于x的六次减去1，则结果为7

Answer 4

(x+1)(x-1)(x²+x+1)(x²-x+1)
=（x²-1）(x²+x+1)(x²-x+1)
=（x²-1)〔 (x²+1)+x 〕〔 (x²+1)-x 〕
=（x²-1)〔( x²+1)²- x²〕
=（x²-1)( x^4+1+x²)
=7

Answer 5

（x+1）（x^2-x-1）=x^3-2*x-1
(x-1)(x²+x+1)=x^然后再相乘