Câu hỏi của Hồ Châu Thái lam

Question

giải phương trình

a) $\frac{4x-17}{2x^2+5}=0$

b) $\frac{\left(x^2-2x\right)-\left(3x+6\right)}{x+2}=0$

c) $\frac{x^2-x-6}{x-3}=0$

d) $\frac{2x-5}{x+5}=3$

e) $\frac{5}{3x+2}=2x-1$

Nguyễn Lê Phước Thịnh · Accepted Answer

a)Ta có: $\frac{4x-17}{2x^2+5}=0$

$\Leftrightarrow4x-17=0$

$\Leftrightarrow4x=17$

$\Leftrightarrow x=\frac{17}{4}$

Vậy: $x=\frac{17}{4}$

b) ĐKXĐ: x≠-2

Ta có: $\frac{\left(x^2-2x\right)-\left(3x+6\right)}{x+2}=0$

$\Leftrightarrow x^2-2x-3x-6=0$

$\Leftrightarrow x^2-5x-6=0$

$\Leftrightarrow x^2+x-6x-6=0$

$\Leftrightarrow x\left(x+1\right)-6\left(x+1\right)=0$

$\Leftrightarrow\left(x+1\right)\left(x-6\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x+1=0\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\x=6\end{matrix}\right.$(tm)

Vậy: x∈{-1;6}

c) ĐKXĐ: x≠3

Ta có: $\frac{x^2-x-6}{x-3}=0$

$\Leftrightarrow x^2-x-6=0$

$\Leftrightarrow x^2-3x+2x-6=0$

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

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

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

Vậy: x=-2

d) ĐKXĐ: x≠-5

Ta có: $\frac{2x-5}{x+5}=3$

⇔$\frac{2x-5}{x+5}-3=0$

⇔$\frac{2x-5}{x+5}-\frac{3\left(x+5\right)}{x+5}=0$

$\Leftrightarrow2x-5-3\left(x+5\right)=0$

$\Leftrightarrow2x-5-3x-15=0$

$\Leftrightarrow-x-20=0$

$\Leftrightarrow-\left(x+20\right)=0$

$\Leftrightarrow x=-20$(tm)

Vậy: x=-20Câu trả lời: a)Ta có: $\frac{4x-17}{2x^2+5}=0$