4) |3 - 2x| = x + 2
5) |2x - 1| = 5 - x
6) |- 3x| = x - 2
7) |2 - 3x| = 2x + 1
8) |2x - 1| + |4x ^ 2 - 1| = 0
9) (2x + 5)/(x + 3) + 1 = 4/(x ^ 2 + 2x - 3) - (3x - 1)/(1 - x)
10) (x - 1)/(x + 3) - x/(x - 3) = (7x - 3)/(9 - x ^ 2)
11) 5 + 96/(x ^ 2 - 16) = (2x - 1)/(x + 4) + (3x - 1)/(x - 4)
12) (2x)/(2x - 1) + x/(2x + 1) = 1 + 4/((2x - 1)(2x + 1))
13) (x + 2)/(x - 2) - 1/x = 2/(x ^ 2 - 2x)
14) x/(2x - 6) + x/(2x + 2) = (2x + 4)/(x ^ 2 - 2x - 3)
9) Ta có: \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)+x^2+2x-3=4+\left(3x-1\right)\left(x+3\right)\)
\(\Leftrightarrow2x^2-2x+5x-5+x^2+2x-3-4-3x^2-10x+x+3=0\)
\(\Leftrightarrow-4x=9\)
hay \(x=-\dfrac{9}{4}\)
10) Ta có: \(\dfrac{x-1}{x+3}-\dfrac{x}{x-3}=\dfrac{7x-3}{9-x^2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3-7x}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(x^2-4x+3-x^2-3x-3+7x=0\)
\(\Leftrightarrow0x=0\)(luôn đúng)
Vậy: S={x|\(x\notin\left\{3;-3\right\}\)}
11) Ta có: \(\dfrac{5+9x}{x^2-16}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}\)
\(\Leftrightarrow\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{9x+5}{\left(x-4\right)\left(x+5\right)}\)
Suy ra: \(2x^2-9x+4+3x^2+12x-x-4-9x-5=0\)
\(\Leftrightarrow5x^2-7x=0\)
\(\Leftrightarrow x\left(5x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{7}{5}\end{matrix}\right.\)
12) Ta có: \(\dfrac{2x}{2x-1}+\dfrac{x}{2x+1}=1+\dfrac{4}{\left(2x-1\right)\left(2x+1\right)}\)
\(\Leftrightarrow\dfrac{2x\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{x\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{4x^2-1+4}{\left(2x-1\right)\left(2x+1\right)}\)
Suy ra: \(4x^2+2x+2x^2-x-4x^2-3=0\)
\(\Leftrightarrow2x^2+x-3=0\)
\(\Leftrightarrow2x^2+3x-2x-3=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\)
13) Ta có: \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\)
\(\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)
Suy ra: \(x^2+2x-x+2-2=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)
4) Ta có: \(\left|3-2x\right|=x+2\)
\(\Leftrightarrow\left[{}\begin{matrix}3-2x=x+2\left(x\le\dfrac{3}{2}\right)\\2x-3=x+2\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x-x=2-3\\2x-x=2+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x=-1\\x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\left(nhận\right)\\x=5\left(nhận\right)\end{matrix}\right.\)
5) Ta có: \(\left|2x-1\right|=5-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5-x\left(x\ge\dfrac{1}{2}\right)\\2x-1=x-5\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x+x=5+1\\2x-x=-5+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=6\\x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
6) Ta có: \(\left|-3x\right|=x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x=x-2\left(x\le0\right)\\3x=x-2\left(x>0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x-x=-2\\3x-x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-4x=-2\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(loại\right)\\x=-1\left(loại\right)\end{matrix}\right.\)
7) Ta có: \(\left|2-3x\right|=2x+1\)
\(\Leftrightarrow\left[{}\begin{matrix}2-3x=2x+1\left(x\le\dfrac{2}{3}\right)\\3x-2=2x+1\left(x>\dfrac{2}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x-2x=1-2\\3x-2x=1+2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-5x=-1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)
