[x-y/2=2+x/3=y+z/4,x+y+z=27]

来源：百度知道编辑：UC知道时间：2024/09/21 08:06:25

令x+y=2k z+x=3k y+z=4k
则x+y+z=9k=27
k=3

带入x+y=6 z+x=9 y+z=12
x=1.5
y=4.5
z=7.5

首先把第一个方程去分母,得到6x-6y=24+4x=12y+3z 此式可化为三个方程,
6x-6y=24+4x,6x-6y=12y+3z,24+4x=12y+3z 再简化,x=12+3y,z=2x-6y,第三个方程可以不要.接着把x=12+3y代入z=2x-6y,得z=24,
x+y+z=x+y+24=12+3y+y+24=27
y=-9/4,
x=21/4

若3x-2y=0,求(x+y)/(x-y)+(x-y)/(x+y)的值已知(x-y)/(x=y)=2,求(x-y)/(2x+2y)-(2x+2y)/(3x-3y)的值已知：X*X+Y*Y=34，X-Y=2，则X/Y= z=根号（(x*x+y*y-x）/(2*x-x*x-y*y)) 求它的定义域若 x/3y = y/2x-5y =6x-15y/x x^2-3y^2=2xy, x+2y/x-y=? (x+y)/3=(x-y)/4 2x+3y=11 1/4(x+y)+1/2(x+y)*(x+y)>=x*根号y+y*根号x 已知|Y/X|=3 求 [（X^2+2XY+Y^2）/X^2Y^2][(x-y)/(x-2y)] x^2+√（2）y=√3, y^2+√（2）x=√3,x≠y,x/y+y/x=?????