Question

[Phan Lê Hoàn Nguyên]

Tìm x,y,z thuộc Q, biết : 
a)  |x+\(\frac{1}{2}\)|+ |y-\(\frac{3}{4}\)|+ |z-1|=0
b)  |x-\(\frac{3}{4}\)|+ |\(\frac{2}{5}\)-y|+ |x-y+z|=0
c) |x-\(\frac{2}{3}\)|+ |x+y+\(\frac{3}{4}\)|+ |y-z-\(\frac{5}{6}\)|=0
d)  |x-\(\frac{1}{2}\)|+ |xy-\(\frac{3}{4}\)|+ |2x-3y-z|=0
e)  |x-\(\frac{2}{3}\)|+ |xy-\(\frac{5}{8}\)|+ |yz+\(\frac{3}{4}\)|≤ 0
f) |xy+\(\frac{2}{3}\)|+|yz-\(\frac{8}{9}\)|+|zx+\(\frac{3}{4}\)|=0

Answer 1

[Thanh Ngân]

\(\left|x+\frac{1}{2}\right|+\left|y-\frac{3}{4}\right|+\left|z-1\right|=0\) \(0\)

<=> \(\hept{\begin{cases}x+\frac{1}{2}=0\\y-\frac{3}{4}=0\\z-1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{3}{4}\\z=1\end{cases}}\)

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

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

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

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

\(\left|x-\frac{2}{3}\right|+\left|x+y+\frac{3}{4}\right|+\left|y-z-\frac{5}{6}\right|=0\)

<=> \(\hept{\begin{cases}x-\frac{2}{3}=0\\x+y+\frac{3}{4}=0\\y-z-\frac{5}{6}=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{2}{3}\\y=\frac{-17}{12}\\z=\frac{-9}{4}\end{cases}}\)

\(\left|x-\frac{1}{2}\right|+\left|xy-\frac{3}{4}\right|+\left|2x-3y-z\right|=0\)

<=> \(\hept{\begin{cases}x-\frac{1}{2}=0\\xy-\frac{3}{4}=0\\2x-3y-z=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{3}{4}:\frac{1}{2}=\frac{3}{2}\\z=\frac{-7}{2}\end{cases}}\)

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