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

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/02 21:39:28
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
xRJ0A2i-T?2 E7nŀqӍ+a6.FA׈ѕ`sHg ]){sr2kw&(L./:fr~rgw`q=e\f\Gg2K&ZIA.)bqvUB\<_ p [ Q#m2 i6nU3@":,6v3 qƒ_+Q.sB C-ztC%̿GvVWY# GX(fjMcXڝp&!XYOQМ^Og& ux#})&w|D b' 86H>ӲӇEy5uS

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&...

全部展开

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

收起