已知x=√2=1分之1,y=√3+1分之2,求x²+2x-2y-y²分之x²y+y-xy&
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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
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)
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)
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