Đặt \(A=\left|x-\frac{1}{2}\right|+\left|y+\frac{2}{3}\right|+\left|x^2+xz\right|\)
Vì \(\left|x-\frac{1}{2}\right|\ge0,\left|y+\frac{2}{3}\right|\ge0,\left|x^2+xz\right|\ge0\Rightarrow\left|x-\frac{1}{2}\right|+\left|y+\frac{2}{3}\right|+\left|x^2+xz\right|\ge0\)
Mà VP=0
\(\Rightarrow A=0\Leftrightarrow\left|x-\frac{1}{2}\right|=0\Leftrightarrow x=\frac{1}{2},\left|y+\frac{2}{3}\right|=0\Leftrightarrow y=-\frac{2}{3}\)
\(\Leftrightarrow\left|\left(\frac{1}{2}\right)^2+\frac{1}{2}z\right|=0\Leftrightarrow\left|\frac{1}{4}+\frac{1}{2}z=0\right|\Leftrightarrow\frac{1}{2}z=-\frac{1}{4}\Leftrightarrow z=-\frac{1}{2}\)
Vậy \(x=\frac{1}{2},y=-\frac{2}{3},z=-\frac{1}{2}\)
ta có
x-1/2=0
x=1/2
ta có
y+2/3=0
y=-2/3
ta có: x^2+xz=0
thay số:(1/2)^2+1/2*z=0
1/4+1/2*z=0
1/2*z=-1/4
z=-1/4:1/2
z=1/2
Vậy x=1/2 ;y=-2/3; z=1/2
