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Ta có
\(\begin{cases}\left|x-\frac{1}{2}\right|\ge0\\\left|y+\frac{3}{2}\right|\ge0\\\left|x+y-z-\frac{1}{2}\right|\ge0\end{cases}\)
Maf \(\left|x-\frac{1}{2}\right|+\left|y+\frac{3}{2}\right|+\left|x+y-z-\frac{1}{2}\right|=0\)
\(\Rightarrow\begin{cases}x-\frac{1}{2}=0\\y+\frac{3}{2}=0\\x+y-z-\frac{1}{2}=0\end{cases}\)
\(\Rightarrow\begin{cases}x=\frac{1}{2}\\y=-\frac{3}{2}\\x+y-z=\frac{1}{2}\end{cases}\)
\(\Rightarrow\begin{cases}x=\frac{1}{2}\\y=-\frac{3}{2}\\\frac{1}{2}-\frac{3}{2}-z=\frac{1}{2}\end{cases}\)
\(\Rightarrow\begin{cases}x=\frac{1}{2}\\y=-\frac{3}{2}\\-z=\frac{3}{2}\end{cases}\)
\(\Rightarrow\begin{cases}x=\frac{1}{2}\\y=-\frac{3}{2}\\z=-\frac{3}{2}\end{cases}\)
Giải:
Vì:
\(\left\{{}\begin{matrix}\left|3x-\dfrac{1}{2}\right|\ge0\\\left|\dfrac{1}{2}y+\dfrac{3}{5}\right|\ge0\end{matrix}\right.\)
Nên dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\left|3x-\dfrac{1}{2}\right|=0\\\left|\dfrac{1}{2}y+\dfrac{3}{5}\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{2}y+\dfrac{3}{5}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=\dfrac{1}{2}\\\dfrac{1}{2}y=-\dfrac{3}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{6}{5}\end{matrix}\right.\)
Vậy ...
b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{1}{5}y-\dfrac{1}{2}\right|\le0\)
Vì:
\(\left\{{}\begin{matrix}\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|\ge0\\\left|\dfrac{1}{5}y-\dfrac{1}{2}\right|\ge0\end{matrix}\right.\)
Dấu "=" xảy ra, khi và chỉ khi:
\(\left\{{}\begin{matrix}\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|=0\\\left|\dfrac{1}{5}y-\dfrac{1}{2}\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{1}{5}y-\dfrac{1}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x=-\dfrac{1}{9}\\\dfrac{1}{5}y=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{5}{2}\end{matrix}\right.\)
Vậy ...
a) \(\left|3x-\frac{1}{2}\right|+\left|\frac{1}{2}y+\frac{3}{5}\right|=0\)
\(\Rightarrow\left|3x-\frac{1}{2}\right|=0\) \(\Rightarrow\left|\frac{1}{2}y+\frac{3}{5}\right|=0\)
\(\Rightarrow3x-\frac{1}{2}=0\) \(\Rightarrow\frac{1}{2}y+\frac{3}{5}=0\)
\(3x=\frac{1}{2}\) \(\frac{1}{2}y=\frac{-3}{5}\)
\(x=\frac{1}{2}:3\) \(y=\left(\frac{-3}{5}\right):\frac{1}{2}\)
\(x=\frac{1}{6}\) \(y=\frac{-6}{5}\)
KL: x = 1/6; y = -6/5
b) \(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\le0\)
mà \(\left|\frac{3}{2}x+\frac{1}{9}\right|>0;\left|\frac{1}{5}y-\frac{1}{2}\right|>0\)
\(\Rightarrow\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|>0\)
=> trường hợp |3/2x +1/9| + |1/5y -1/2| < 0 không thế xảy ra
\(\Rightarrow\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|=0\)
rùi bn lm tương tự như phần a nhé!