试比较(x²-√2x+1)(x²+√2x+1)与(x²-x+1)(x²+x+1)大小

试比较(x²-√2x+1)(x²+√2x+1)与(x²-x+1)(x²+x+1)大小
试比较(x²-√2x+1)(x²+√2x+1)与(x²-x+1)(x²+x+1)大小

试比较(x²-√2x+1)(x²+√2x+1)与(x²-x+1)(x²+x+1)大小
把x²+1和√2x看成整体
形如(a+b)(a-b) = a² - b²
就可以顺利得出(x²+1)²-(√2x)²

(x²-√2x+1)(x²+√2x+1)
=((x²+1)-√2x) * ((x²+1)+√2x)
=(x²+1)²-（√2x)²+√2x*（x²+1）-√2x*（x²+1）
=(x²+1)²-（√2x)²

这步应用到平方差公式：（a+b）（a-b）=a²-b²
(x²-√2x+1)(x²+√2x+1)
=【(x²+1)-(√2x)】【(x²+1)+(√2x)】
=(x²+1)²-(√2x)²