Câu hỏi của Sách Giáo Khoa

Question

Giải các phương trình :

a. $x+1+\dfrac{2}{x+3}=\dfrac{x+5}{x+3}$

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

c. $\dfrac{x^2-4x-2}{\sqrt{x-2}}=\sqrt{x-2}$

d. \(\dfrac{2x^2-x-3}{\sqrt{2x-3}}=\sqrt{...

Kuro Kazuya · Accepted Answer

a) $x+1+\dfrac{2}{x+3}=\dfrac{x+5}{x+3}$

$\Leftrightarrow x+\dfrac{x+5}{x+3}=\dfrac{x+5}{x+3}$

$\Leftrightarrow x=0$

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

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

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

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

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

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

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

$\Leftrightarrow x^2-x+3=3x-x^2+x$ ( điều kiện $x\ne1$ )

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

$\Delta=b^2-4ac$

$\Delta=1$

$\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3}{2}\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=1\left(loại\right)\end{matrix}\right.$

Vậy $x=\dfrac{3}{2}$

c) $\dfrac{x^2-4x-2}{\sqrt{x-2}}=\sqrt{x-2}$

$\Leftrightarrow x^2-4x-2=\sqrt{\left(x-2\right)^2}$ ( điều kiện $x>2$ )

$\Leftrightarrow x^2-4x-2=x-2$

$\Leftrightarrow x^2-5x=0$

$\Leftrightarrow x\left(x-5\right)=0$

$\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\x=5\end{matrix}\right.$

Vậy $x=5$

d) $\dfrac{2x^2-x-3}{\sqrt{2x-3}}=\sqrt{2x-3}$

$\Leftrightarrow2x^2-x-3=\sqrt{\left(2x-3\right)^2}$ ( điều kiện $x>\dfrac{3}{2}$ )

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

$\Leftrightarrow2x^2-3x=0$

$\Leftrightarrow x\left(2x-3\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x=0\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\x=\dfrac{3}{2}\left(loại\right)\end{matrix}\right.$

Vậy phương trình vô nghiệmCâu trả lời: a) $x+1+\dfrac{2}{x+3}=\dfrac{x+5}{x+3}$