作业帮已知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
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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
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
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
x=1/(√2-1)=√2+1
y=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+2)
=(√6+√3-√2-2)/(√3+√2+2)
答：
x=1/(√2+1)=(√2-1)/[(√2-1)(√2+1)]=√2-1
y=2/(√3+1)=2(√3-1)/[(√3-1)(√3+1)]=√3-1
xy=(√2-1)(√3-1)=√6-√2-√3+1
x²+2x-2y-y²分之x²y+y-xy²-x
=(x²y+y-xy²-x)/(x&...
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答：
x=1/(√2+1)=(√2-1)/[(√2-1)(√2+1)]=√2-1
y=2/(√3+1)=2(√3-1)/[(√3-1)(√3+1)]=√3-1
xy=(√2-1)(√3-1)=√6-√2-√3+1
x²+2x-2y-y²分之x²y+y-xy²-x
=(x²y+y-xy²-x)/(x²+2x-2y-y²)
=[(x-y)xy-(x-y)]/[(x-y)(x+y)+2(x-y)]
=(xy-1)/(x+y+2)
=(√6-√2-√3)/(√3+√2)
=(√6-√2-√3)(√3-√2)
=3√2-2√3-√6+2-3+√6
=3√2-2√3-1
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