先化简,后求值:(x^2y-4y^3)/(x^2+4xy+4y^3)*((4xy)/(x-2y)+x)其中x=根号2-1y=根号2+1

来源:学生作业帮助网 编辑:作业帮 时间:2024/07/14 11:11:08
先化简,后求值:(x^2y-4y^3)/(x^2+4xy+4y^3)*((4xy)/(x-2y)+x)其中x=根号2-1y=根号2+1
x){gu :O'=a=4**uM*5AlmJm0WKCUjjWh>md g v>nkX cj$;zf۟jy:a9.Ӎ0My2N&jʆ tOmF6P!D>U2JA5B 6 ECE%~ԱQ,Wt*,OulntqAb4m{9!@mT%J

先化简,后求值:(x^2y-4y^3)/(x^2+4xy+4y^3)*((4xy)/(x-2y)+x)其中x=根号2-1y=根号2+1
先化简,后求值:(x^2y-4y^3)/(x^2+4xy+4y^3)*((4xy)/(x-2y)+x)其中x=根号2-1y=根号2+1

先化简,后求值:(x^2y-4y^3)/(x^2+4xy+4y^3)*((4xy)/(x-2y)+x)其中x=根号2-1y=根号2+1
题是这样的吧(x^2y-4y^3)/(x^2+4xy+4y^2)*((4xy)/(x-2y)+x),感觉是4y²
原式=y(x+2y)(x-2y)/(x+2y)² * (4xy +x²-2xy)/(x-2y)
=y(x+2y)(x-2y)/(x+2y)² * x(x+2y)/(x-2y)
=xy
∵x=√2-1 ,y=√2+1
∴xy=1
∴原式=1

题是这样的吧(x^2y-4y^3)/(x^2+4xy+4y^2)*((4xy)/(x-2y)+x),?感觉是4y²
原式=y(x+2y)(x-2y)/(x+2y)² * (4xy +x²-2xy)/(x-2y)
=y(x+2y)(x-2y)/(x+2y)² * x(x+2y)/(x-2y)
=xy
∵x=√2-1 ,y=√2+1
∴xy=1
∴原式=1