Question

6．已知x=$\frac{1}{\sqrt{2}+1}$，y=$\frac{1}{\sqrt{2}-1}$，求$\frac{{x}^{3}-{x}^{2}y+x{y}^{2}-{y}^{3}}{x-y}$的值．

Answer 1

分析 由x=$\frac{1}{\sqrt{2}+1}$=$\sqrt{2}$-1，y=$\frac{1}{\sqrt{2}-1}$=$\sqrt{2}$+1，進(jìn)一步化簡(jiǎn)分式代入球的答案即可．

解答 解：∵x=$\frac{1}{\sqrt{2}+1}$=$\sqrt{2}$-1，y=$\frac{1}{\sqrt{2}-1}$=$\sqrt{2}$+1，
∴$\frac{{x}^{3}-{x}^{2}y+x{y}^{2}-{y}^{3}}{x-y}$
=$\frac{{x}^{2}（x-y）+{y}^{2}（x-y）}{x-y}$
=x²+y²
=3-2$\sqrt{2}$+3+2$\sqrt{2}$
=6．

點(diǎn)評(píng) 此題考查二次根式的化簡(jiǎn)求值，注意先化簡(jiǎn)二次根式和分式，再進(jìn)一步代入求得答案即可．