Question

tìm các số hữu tỉ x, y, z biết:

x(x+ y+ z)= -5 , y(x+ y+ z)=9, z(x+ y+z)=5

Answer 1

Ta có :

\(x\left(x+y+z\right)=-5\)

\(y\left(x+y+z\right)=9\)

\(z\left(x+y+z\right)=-5\)

\(\Rightarrow x\left(x+y+z\right)+y\left(x+y+z\right)+z\left(x+y+z\right)=-5+9+-5\)

\(\Rightarrow\left(x+y+z\right)\left(x+y+z\right)=9\)

\(\Rightarrow\left(x+y+z\right)^2=3^2=\left(-3\right)^2\)

Với \(\left(x+y+z\right)=3\); ta có:

\(x=-5:\left(x+y+z\right)=-5:3=-\frac{5}{3}\)

\(y=9:\left(x+y+z\right)=9:3=3\)

\(z=5:\left(x+y+z\right)=5:3=\frac{5}{3}\)

Với \(\left(x+y+z\right)=-3\)

\(x=-5:\left(x+y+z\right)=-5:\left(-3\right)=\frac{5}{3}\)

\(y=9:\left(x+y+z\right)=9:\left(-3\right)=-3\)

\(z=5:\left(x+y+z\right)=5:\left(-3\right)=-\frac{5}{3}\)

Answer 2

x(x+y+z) + y(x+y+z) + z(x+y+z) = (-5) + 9 + 5   
suy ra (x+y+z ) ( x+y+z ) = 9
          (x+y+z)^2 = 9 
x+y+z = -3 hoặc 3 
đến đây thay vào đề bài là làm được