Question

[Nguyễn Hằng]

TÌM x, y, z thuộc Q:

b, I x-3/4I+I 2/5-yI+Ix-y+zI=0

Answer 1

[Xyz OLM]

Ta có : \(\hept{\begin{cases}\left|x-\frac{3}{4}\right|\ge0\forall x\\\left|\frac{2}{5}-y\right|\ge0\forall y\\\left|x-y+z\right|\ge0\forall x;y;z\end{cases}}\Leftrightarrow\left|x-\frac{3}{4}\right|+\left|\frac{2}{5}-y\right|+\left|x-y+z\right|\ge0\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-\frac{3}{4}=0\\\frac{2}{5}-y=0\\x-y+z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=\frac{2}{5}\\z=-\frac{7}{20}\end{cases}}\)

Vậy x = 3/4 ; y = 2/5 ; z = -7/20 

Answer 2

[Yen Nhi]

\(\left|x-\frac{3}{4}\right|+\left|\frac{2}{5}-y\right|+\left|x-y+z\right|=0\)

Ta có: \(\left|x-\frac{3}{4}\right|;\left|\frac{2}{5}-y\right|;\left|x-y+z\right|\ge0\Rightarrow\left|x-\frac{3}{4}\right|+\left|\frac{2}{5}-y\right|+\left|x-y+z\right|\ge0\)

Mà \(\left|x-\frac{3}{4}\right|+\left|\frac{2}{5}-y\right|+\left|x-y+z\right|=0\)

\(\Rightarrow\hept{\begin{cases}x-\frac{3}{4}=0\\\frac{2}{5}-y=0\\x-y+z=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=\frac{2}{5}\\\frac{3}{4}-\frac{2}{5}+z=0\Rightarrow z=\frac{-7}{20}\end{cases}}\)