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\(\Leftrightarrow\left(2x-1\right)\left[\left(5x-3\right)-\left(2x-1\right)\right]=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)
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\(\dfrac{x+3}{x}=\dfrac{2x+2}{2x-1}\) (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 \(S=\left\{1\right\}\)
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a: \(\Leftrightarrow30\left(x-3\right)-16=9\left(x-1\right)+72\)
\(\Leftrightarrow30x-90-16=9x-9+72\)
=>30x-106=9x+63
=>21x=169
hay x=169/21
b: =>4x+20=2x-3
=>2x=-23
hay x=-23/2
3 - 2x = 3(x + 1) - x - 2
⇔ 3 - 2x = 3x + 3 - x - 2
⇔ 4x = 2
⇔ x = 1/2