(x+2y-z)(x-2y+z)
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(x+2y-z)(x-2y+z)(x+2y-z)(x-2y+z) (x+2y-z)(x-2y+z)稍等=(x+(2y–z))(x–(2y–z))=x2–(2y–z)2=x2–4y2+4yz–
(x+2y-z)(x-2y+z)
(x+2y-z)(x-2y+z)
(x+2y-z)(x-2y+z)
稍等
=(x+(2y–z))(x–(2y–z))=x2–(2y–z)2=x2–4y2+4yz–z2
(x-2y+z)(-x+2y+z)
x+2y-z)(x+2y-z)
(x+2y-z)(x-2y+z)
化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(x-2z+y)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z)
计算:(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z)
计算:x^2/(x-y)(x-z)+y^2/(y-x)(y-z)+z^2/(z-x)(z-y)
化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(xy-2z)(y+z-2x)+(x-z)(y-z)/(y+z-2x)(x-2y+z)速速回答
化简(x+Y)^2(y+z-x)(z+x-y)+(x-y)^2(x+y+z)(x+y-z)
计算题(x+y)^2(y+z-x)(z+x-y)+(x-y)^2(x+y+z)(x+y-z).
1、y-x/x²-y²2、(x-y)(y-z)(z-x)/(z-y)(y-x)(x-z)
x,y,z为正实数 x/(2x+y+z)+y/(x+2y+z)+z/(x+y+2z)
用行列式的性质证明:y+z z+x x+y x y z x+y y+z z+x =2 z x y z+x x+y y+z y z x 这个怎么证?
化简:(2x-y-z)/(x-y)(x-z)+(2y-z-x)/(y-z)(y-x)+(2z-x-y)/(z-x)(z-y)化简:(2x-y-z)/(x-y)(x-z)+(2y-z-x)/(y-z)(y-x)+(2z-x-y)/(z-x)(z-y)我知道把分母化成一样的、但是之后怎么算?
x,y,z正整数 x>y>z证明 x^2x +y^2y+z^2z>x^(y+z)*y^(x+z)*z^(x+y)x,y,z正整数 x>y>z证明 x^2x * y^2y * z^2z>x^(y+z)*y^(x+z)*z^(x+y)不是+是 *
(x-2y+z)(x+y-2z)分之(y-x)(z-x) + (x+y-2z)(y+z-2x)分之(z-y)(x-y) + (y+z-2z)(x-2y+z)分之(x-z)(y-z)=?第三部分那个是 (y+z-2x)(x-2y+z)分之(x-z)(y-z)
试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
化简(y-z)^2/(x-y)(x-z)+(z-x)^2/(y-x)(y-z)+(x-y)^2/(z-x)(z-y)
已知:x^2/(z+y)+y^2/(x+z)+z^2/(x+y)=0,求x/(z+y)+y/(x+z)+z/(x+y)的值.