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\(\Leftrightarrow\left(2x-1\right)\left[\left(5x-3\right)-\left(2x-1\right)\right]=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)
\(\dfrac{x+3}{x}=\dfrac{2x+2}{2x-1}\) (ĐKXĐ: \(x\ne0;x\ne\dfrac{1}{2}\))
\(\)\(\Leftrightarrow\dfrac{x+3}{x}=\dfrac{2\left(x+1\right)}{2x-1}\Leftrightarrow\left(x+3\right)\left(2x-1\right)=2x\left(x+1\right)\)
\(\Leftrightarrow2x^2+6x-x-3=2x^2+2x\)
\(\Leftrightarrow2x^2-2x^2+6x-x-2x=3\)
\(\Leftrightarrow3x=3\Leftrightarrow x=1\left(TM\right)\)
\(\Rightarrow S=\left\{1\right\}\)
\(\dfrac{x+3}{x}=\dfrac{2x+2}{2x-1}\)
\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)=x\left(2x+2\right)\)
\(\Leftrightarrow2x^2-x+6x-3=2x^2+2x\)
\(\Leftrightarrow2x^2+5x-3-2x^2-2x=0\)
\(\Leftrightarrow3x-3=0\)
\(\Leftrightarrow3\left(x-1\right)=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy \(S=\left\{1\right\}\)
ĐKXĐ: \(x\ne3;x\ne-5\)
\(\frac{2\left|x-3\right|}{x^2+2x-15}=1\Rightarrow2\left|x-3\right|=x^2+2x-15\)
\(\Rightarrow2\left|x-3\right|=\left(x-3\right)\left(x+5\right)\Leftrightarrow\left(x-3\right)\left(x+5\right)-2\left|x-3\right|=0\)
Với \(x>3\) thì \(\Leftrightarrow\left(x-3\right)\left(x+5\right)-2\left(x-3\right)=0\Leftrightarrow\left(x+3\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\left(KTM\right)\)
Với \(x< 3\) thì: \(\Leftrightarrow\left(x-3\right)\left(x+5\right)+2\left(x-3\right)=0\Leftrightarrow\left(x-3\right)\left(x+7\right)=0\Rightarrow\left[{}\begin{matrix}x=-7\left(TM\right)\\x=3\left(KTM\right)\end{matrix}\right.\)
pt có nghiệm duy nhất x=-7
3 - 2x = 3(x + 1) - x - 2
⇔ 3 - 2x = 3x + 3 - x - 2
⇔ 4x = 2
⇔ x = 1/2