a/ Ta có :
\(\left|x-2\right|=\left|2-x\right|\)
\(\Leftrightarrow\left|x\right|+\left|x-2\right|=\left|x\right|+\left|2-x\right|\)
\(\Leftrightarrow\left|x\right|+\left|x-2\right|\ge\left|x+2-x\right|\)
\(\Leftrightarrow\left|x\right|+\left|x-2\right|\ge2\)
Dấu "=" xảy ra khi :
\(x\left(x-2\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x-2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\x-2\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x\ge2\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\x\le2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow2\le x\le2\Leftrightarrow x=2\)
Vậy \(x=2\)
b/ \(\left|2x-1\right|+\left|9-2x\right|\ge\left|2x-1+9-2x\right|\)
\(\Leftrightarrow\left|2x-1\right|+\left|9-2x\right|\ge8\)
Dấu "=" xảy ra khi :
\(\left(2x-1\right)\left(9-2x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-1\ge0\\9-2x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1\le0\\9-2x\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\le\dfrac{9}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\x\ge\dfrac{9}{2}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\dfrac{1}{2}\le x\le\dfrac{9}{2}\)
Vậy ....
c/ Ta có : \(\left|3x-20\right|=\left|20-3x\right|\)
\(\Leftrightarrow\left|3x+7\right|+\left|3x-20\right|=\left|3x+7\right|+\left|20-3x\right|\)
\(\Leftrightarrow\left|3x+7\right|+\left|3x-20\right|\ge\left|3x+7+20-3x\right|\)
\(\Leftrightarrow\left|3x+7\right|+\left|3x-20\right|\ge27\)
Dấu "=" xảy ra khi :
\(\left(3x+7\right)\left(20-3x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x+7\ge0\\20-3x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}3x+7\le0\\20-3x\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-\dfrac{7}{3}\\x\le\dfrac{20}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-\dfrac{7}{3}\\x\ge\dfrac{20}{3}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\dfrac{20}{3}\le x\le-\dfrac{7}{3}\)
Vậy...
d/ \(\left|10-x\right|+\left|x+30\right|\ge\left|10-x+x+30\right|\)
\(\Leftrightarrow\left|10-x\right|+\left|x+30\right|\ge40\)
Dấu "=" xảy ra khi :
\(\left(10-x\right)\left(x+30\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}10-x\ge0\\x+30\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}10-x\le0\\x+30\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}10\ge x\\x\ge-30\end{matrix}\right.\\\left\{{}\begin{matrix}10\le x\\x\le-30\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow10\ge x\ge-30\)
Vậy...
