7) \(\left|2-3x\right|=2x+1\)
Có \(\left|2-3x\right|=\) \(2-3x\) nếu \(2-3x\ge0\Leftrightarrow x\le\dfrac{2}{3}\)
\(\left|2-3x\right|=\) 3x-2 nếu \(2-3x< 0\Leftrightarrow x>\dfrac{2}{3}\)
Trường hợp 1: nếu \(\left|2-3x\right|=\) \(2-3x\) thì ta được:
\(2-3x=2x+1\Leftrightarrow1=5x\Leftrightarrow x=\dfrac{1}{5}\)
Trường hợp 2 nếu \(\left|2-3x\right|=\) 3x-2 thì ta được:
\(3x-2=2x+1\Leftrightarrow3=x\)
Vậy \(x\in\left\{\dfrac{1}{5};3\right\}\)
8) \(\left|2x-1\right|+\left|4x^2-1\right|=0\)
\(\Leftrightarrow\left|2x-1\right|+\left|\left(2x-1\right).\left(2x+1\right)\right|=0\)
\(\Leftrightarrow\left|2x-1\right|+\left|2x-1\right|.\left|2x+1\right|=0\)
\(\Leftrightarrow\left|2x-1\right|.\left(1+\left|2x+1\right|\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\left|2x-1\right|=0\\1+\left|2x+1\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\\left|2x+1\right|=-1\end{matrix}\right.\) \(\Rightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\)
Vậy x=\(\dfrac{1}{2}\)
9) \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2-x+3x-3}-\dfrac{2x-1}{1-x}\)
\(\Leftrightarrow\dfrac{2x+5+x+3}{x+3}=\dfrac{4}{x.\left(x-1\right)+3.\left(x-1\right)}+\dfrac{2x-1}{x-1}\)
\(\Leftrightarrow\dfrac{3x+8}{x+3}=\dfrac{4}{\left(x+3\right).\left(x-1\right)}+\dfrac{2x-1}{x-1}\)
\(\Leftrightarrow\dfrac{\left(3x+8\right).\left(x-1\right)}{\left(x+3\right).\left(x-1\right)}=\dfrac{4+\left(2x-1\right).\left(x+3\right)}{\left(x+3\right).\left(x-1\right)}\)
\(\Rightarrow\left(3x+8\right).\left(x-1\right)=4+\left(2x-1\right).\left(x+3\right)\)
\(\Leftrightarrow3x^2-3x+8x-8=4+2x^2+6x-x-3\)
\(\Leftrightarrow3x^2+5x-8=2x^2+5x+1\)
\(\Leftrightarrow x^2-9=0\)
\(\Leftrightarrow\left(x-3\right).\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Vậy x\(\in\left\{-3;3\right\}\)
10)\(\dfrac{x-1}{x+3}-\dfrac{x}{x-3}=\dfrac{7x-3}{\left(3-x\right).\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{\left(x-1\right).\left(3-x\right)}{\left(x+3\right).\left(3-x\right)}+\dfrac{x.\left(x+3\right)}{\left(3-x\right).\left(x+3\right)}=\dfrac{7x-3}{\left(x+3\right).\left(3-x\right)}\)
\(\Leftrightarrow\dfrac{\left(x-1\right).\left(3-x\right)+x.\left(x+3\right)}{\left(x+3\right).\left(3-x\right)}=\dfrac{7x-3}{\left(x+3\right).\left(3-x\right)}\)
\(\Rightarrow3x-x^2-3+x+x^2+3x=7x-3\)
\(\Leftrightarrow7x-3=7x-3\) (đúng với mọi x) .
Vậy phương trình vô số nghiệm.
