已知x=√2=1分之1,y=√3+1分之2,求x²+2x-2y-y²分之x²y+y-xy²-x的值

题目

已知x=√2=1分之1,y=√3+1分之2,求x²+2x-2y-y²分之x²y+y-xy²-x的值

答案

x=1/(√2-1)=√2+1y=2/(√3+1)=2(√3-1)/[(√3+1)(√3-1)]=√3-1 (x^2y+y-xy^2-x)/(x^2+2x-2y-y^2)=[xy(x-y)+(y-x)]/[(x^2-y^2)+2(x-y)]=[(x-y)(xy-1)]/[(x-y)(x+y+2)]=(xy-1)/(x+y+2)=[(√3-1)(√2+1)-1]/(√3+√2+...