Câu hỏi của hatsune miku

Question

Giaỉ pt: $\dfrac{x+3}{x}=\dfrac{2x+2}{2x-1}$

Thierry Henry · Accepted Answer

$\dfrac{x+3}{x}=\dfrac{2x+2}{2x-1}$ (ĐKXĐ: $x\ne0;x\ne\dfrac{1}{2}$)

$\Leftrightarrow\dfrac{x+3}{x}=\dfrac{2\left(x+1\right)}{2x-1}\Leftrightarrow\left(x+3\right)\left(2x-1\right)=2x\left(x+1\right)$

$\Leftrightarrow2x^2+6x-x-3=2x^2+2x$

$\Leftrightarrow2x^2-2x^2+6x-x-2x=3$

$\Leftrightarrow3x=3\Leftrightarrow x=1\left(TM\right)$

$\Rightarrow S=\left\{1\right\}$Câu trả lời: $\dfrac{x+3}{x}=\dfrac{2x+2}{2x-1}$ (ĐKXĐ: $x\ne0;x\ne\dfrac{1}{2}$)

$\Leftrightarrow\dfrac{x+3}{x}=\dfrac{2\left(x+1\right)}{2x-1}\Leftrightarrow\left(x+3\right)\left(2x-1\right)=2x\left(x+1\right)$

$\Leftrightarrow2x^2+6x-x-3=2x^2+2x$

$\Leftrightarrow2x^2-2x^2+6x-x-2x=3$

$\Leftrightarrow3x=3\Leftrightarrow x=1\left(TM\right)$

$\Rightarrow S=\left\{1\right\}$

Vorlage (bạn Chi) · Answer

$\dfrac{x+3}{x}=\dfrac{2x+2}{2x-1}$

$\Leftrightarrow\left(x+3\right)\left(2x-1\right)=x\left(2x+2\right)$

$\Leftrightarrow2x^2-x+6x-3=2x^2+2x$

$\Leftrightarrow2x^2+5x-3-2x^2-2x=0$

$\Leftrightarrow3x-3=0$

$\Leftrightarrow3\left(x-1\right)=0$

$\Leftrightarrow x-1=0$

$\Leftrightarrow x=1$

Vậy $S=\left\{1\right\}$Câu trả lời: $\dfrac{x+3}{x}=\dfrac{2x+2}{2x-1}$