Question

Tìm x,y,z

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

Answer 1

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

=> \(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|=0\\\left|y+\dfrac{2}{3}\right|=0\\\left|x^2+xz\right|=0\end{matrix}\right.\)

* \(\left|x-\dfrac{1}{2}\right|\)=0

=> x- \(\dfrac{1}{2}\)=0

=>x= \(\dfrac{1}{2}\)

* \(\left|y+\dfrac{2}{3}\right|\)=0

=> y+\(\dfrac{2}{3}\)=0

=>y = \(\dfrac{-2}{3}\)

* \(\left|x^2+xz\right|\)=0

mà x= \(\dfrac{1}{2}\) (cmt)

=> \(\left(\dfrac{1}{2}\right)^2+\dfrac{1}{2}z=0\)

=>\(\dfrac{1}{2}z=\dfrac{-1}{4}\)

=>z= \(\dfrac{-1}{2}\)

Vậy x,y,z= \(\dfrac{1}{2};\dfrac{-2}{3};\dfrac{-1}{2}\)