Ta có:
x(x+y+z)=-12 , y(x+y+z)=18 , z(x+y+z)=30
=> x(x+y+z) + y(x+y+z) + z(x+y+z) = (-12) +18 + 30
=> (x+y+z)(x+y+z) = 36
=> \(\left(x+y+z\right)^2\)= \(6^2\)
=> x+y+z = 6
Vậy ta có:
x(x+y+z) = 6x = -12
x = -12 : 6 = -2
y(x+y+z) = 6y = 18
y = 18 : 6 = 3
z(x+y+z) = 6z = 30
z = 30 : 6 = 5
Vậy x = -2 ; y = 3 ; z = 5
Tim x, y, z biet x, y, z > 0 , x-y/z = y-z/x = z-x/y va x+2y+3z=30
tim x;y;z biet x/5=y/12=z/7 va x-y+z=30
tim x, y, z, biet 4.x=6.y=12.z va x+y+z=30
Tim x y z biet x/(y-z)=y/(x+z)=z/(y-x)=x-y+z
Tim x,y,z biet x-y+z=x/(y*z)=y/(z+x)=z/(y-x)
tim x,y,z biet; x/z+y+1=y/x+z+1=x+y-2=x+y+z
tim x;y;z biet x(x+y+z)=2; y(x+y+z)=25; z(x+y+z)=-2; x>0
Tim x , y , z biet: x /y+z+1 = y/ z+x+2 = z/ x+y−3 =x+y+z Cach lam ho minh voi
tim x,y,z biet y+z+1=x+z+2/y=x+y-3/2=1/x+y+z
