\(\left|x-\dfrac{1}{3}\right|+\left|x+y\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-\dfrac{1}{3}\right|=0\\\left|x+y\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\x+y=0\Rightarrow y=-\dfrac{1}{3}\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}\left|x-\dfrac{1}{3}\right|\ge0\\\left|x+y\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x-\dfrac{1}{3}\right|+\left|x+y\right|\ge0\)
Đẳng thức xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}\left|x-\dfrac{1}{3}\right|=0\\\left|x+y\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{3}=0\\x+y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\x+y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\\dfrac{1}{3}+y=0\end{matrix}\right.\) \(\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
ahihi cách khác nè @Ái Hân Ngô
\(\left|x-\dfrac{1}{3}\right|+\left|x+y\right|=0\)
Lời giải:
\(\left|x-\dfrac{1}{3}\right|+\left|x+y\right|=\left|\dfrac{1}{3}-x\right|+\left|x+y\right|\)
Áp dụng bđt: \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)
Suy ra: \(\left|\dfrac{1}{3}-x\right|+\left|x+y\right|\ge\left|\dfrac{1}{3}-x+x+y\right|=\left|\dfrac{1}{3}+y\right|\)(1)
Áp dụng bđt: \(\left|a\right|-\left|b\right|\le\left|a-b\right|\)
Suy ra:\(\left|x-\dfrac{1}{3}\right|+\left|x+y\right|\le\left|x-\dfrac{1}{3}-x+y\right|=\left|\dfrac{1}{3}+y\right|\)(2)
Từ \(\left(1\right)+\left(2\right)\) ta có:
\(\left|x-\dfrac{1}{3}\right|+\left|x+y\right|=\left|\dfrac{1}{3}+y\right|=0\)
Suy ra \(y=-\dfrac{1}{3}\)
Từ đó dễ dàng tìm đc \(x=\dfrac{1}{3}\)
