X+Y+Z=26 X-Y=1 2X-Y+Z=18这三个为三元一次方程组,求如何解,要用加减消元法或其他方法!

已知 x/(y+z)+y/(z+x)+z/(x+y)=1求 (x*x)/(y+z)+(y*y)/(x+z)+(z*z)/(x+y)=? X+y+z=26,x-y=1,2x-y+z=18 (x+y-z)(x-y+z)= 分式加减的几道题目求速度1 .(x+y)²/（x-z)(y-z) - (y+z)²/（x-z)(y-z) = 2.2x+3y/x-y+2y/y-x-3y/x-y= 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? 已知实数x y z满足x/(x+1)=y/(y+2)=z/(z+3)=(x+y+z)/3,求x+y+z的值 已知实数x y z满足x/(x+1)=y/(y+2)=z/(z+3)=(x+y+z)/3,求x+y+z的值 已知(x+y)(x+z)=x,(y+z)(y+x)=2y,(z+x)(z+y)=3z,求x,y,z 试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y) 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. 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 1、y-x/x²-y²2、(x-y)(y-z)(z-x)/(z-y)(y-x)(x-z) (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) x+y+z=6 x+2y+3z=14 y+1=z X+2Y= -12X+Y+Z= 1-X+Y-Z= -1 X+Y+Z=? x/y=(x+z)/(y+z)y/z=(x+y)/(x+z) 计算题!【x+y+z=26；x-y=1；2x-y+z=18】!【x/3+y/5；x/2-2y/5=1】