\(a,\)\(\left|x+3\right|=2x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+3=2x+1\\x+3=-\left(2x+1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2x=1-3\\x+3=-2x-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-x=-2\\x+2x=-1-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\3x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(b,\dfrac{x+2}{3}\ge\dfrac{2x-1}{4}-1\)
\(\Leftrightarrow\dfrac{x+2}{3}-\dfrac{2x-1}{4}+1\ge0\)
\(\Leftrightarrow\dfrac{4\left(x+2\right)-3\left(2x-1\right)+12}{12}\ge0\)
\(\Leftrightarrow4x+8-6x+3+12\ge0\)
\(\Leftrightarrow-2x\ge-23\)
\(\Leftrightarrow x\le\dfrac{23}{2}\)
a, \(\left|x+3\right|=2x+1\)
\(\Leftrightarrow\left|x+3\right|\left\{{}\begin{matrix}x+3\\-x-3\end{matrix}\right.\)
\(\left|x+3\right|=x+3\) nếu \(x+3\ge0\Leftrightarrow x\ge-3\)
\(\left|x+3\right|=-x-3\) nếu \(x+3< 0\Leftrightarrow x< -3\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+3=2x+1\left(1\right)\\-x-3=2x+1\left(2\right)\end{matrix}\right.\)
Giải phương trình (1):
\(x+3=2x+1\Leftrightarrow x-2x=1-3\Leftrightarrow-x=-2\Leftrightarrow x=2\) (Thỏa mãn)
Giải phương trình (2):
\(-x-3=2x+1\Leftrightarrow-x-2x=1+3\Leftrightarrow-3x=4\Leftrightarrow x=-\dfrac{4}{3}\) (Ko thỏa mãn)
Vậy phương trình có tập nghiệm là \(S=\left\{2\right\}\)
b, \(\dfrac{x+2}{3}\ge\dfrac{2x-1}{4}-1\)
\(\Leftrightarrow\dfrac{4x+8}{12}\ge\dfrac{6x-3}{12}-\dfrac{12}{12}\)
\(\Leftrightarrow4x+8\ge6x-3-12\)
\(\Leftrightarrow4x-6x\ge-3-12-8\)
\(\Leftrightarrow-2x\ge23\)
\(\Leftrightarrow x\le-\dfrac{23}{2}\)
Vậy bất phương trình có nghiệm là \(x\le-\dfrac{23}{2}\)
