Question

[Nguyễn Minh Hoàng]

Tìm x, y, z:

\(\left|x-\frac{1}{2}\right|+\left|y+\frac{2}{3}\right|+\left|x^2+xz\right|=0\)

Answer 1

[Phùng Minh Quân]

\(\left|x-\frac{1}{2}\right|+\left|y+\frac{2}{3}\right|+\left|x^2+xz\right|=0\)

\(\Leftrightarrow\)\(\left|x-\frac{1}{2}\right|+\left|y+\frac{2}{3}\right|+\left|x\left(x+z\right)\right|=0\)

Ta có : 

\(\left|x-\frac{1}{2}\right|\ge0\)

\(\left|y+\frac{2}{3}\right|\ge0\)

\(\left|x\left(x+z\right)\right|\ge0\)

Mà \(\left|x-\frac{1}{2}\right|+\left|y+\frac{2}{3}\right|+\left|x\left(x+z\right)\right|=0\)

\(\Rightarrow\)\(\hept{\begin{cases}\left|x-\frac{1}{2}\right|=0\\\left|y+\frac{2}{3}\right|=0\\\left|x\left(x+z\right)\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{-2}{3}\\z=\frac{-1}{2}\end{cases}}}\)

Vậy \(x=\frac{1}{2}\)\(;\)\(y=\frac{-2}{3}\) và \(z=\frac{-1}{2}\)

Chúc bạn học tốt ~