Ta có: \(\dfrac{-4x^2}{x^2-25}-\dfrac{2x^2+x}{x^2-25}-\dfrac{2x}{5-x}\)

\(=\dfrac{-4x^2-2x^2-x}{\left(x-5\right)\left(x+5\right)}+\dfrac{2x}{x-5}\)

\(=\dfrac{-6x^2-x}{\left(x-5\right)\left(x+5\right)}+\dfrac{2x\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-6x^2-x+2x^2+10x}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-4x^2+9x}{\left(x-5\right)\left(x+5\right)}\)

B

Đề có sai không bạn?

a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)

\(=\dfrac{3x+2-3x+2-3x+6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)

b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)

\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-5\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{-5}{2x\left(x-5\right)}\)

c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)

\(=\dfrac{-\left(2x-1\right)^2-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-\left(4x^2-4x+1\right)-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-4x^2+4x-1-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)

\(1,\left(dk:x\ne0,-1,4\right)\)

\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)

\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)

\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)

\(\Leftrightarrow-x=-44\)

\(\Leftrightarrow x=44\left(tm\right)\)

\(2,\left(đk:x\ne4\right)\)

\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)

\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)

\(\Leftrightarrow28-12-6x-9+5x-20=0\)

\(\Leftrightarrow-x=13\)

\(\Leftrightarrow x=-13\left(tm\right)\)

Lần sau bạn chú ý ghi đầy đủ yêu cầu của đề.

* Coi đây là bài toán rút gọn

Lời giải:

ĐKXĐ: $x\neq 2$

$\frac{x-5}{x^2-4x+4}:\frac{x^2-25}{2x-4}=\frac{x-5}{(x-2)^2}:\frac{(x-5)(x+5)}{2(x-2)}$

$=\frac{x-5}{(x-2)^2}.\frac{2(x-2)}{(x-5)(x+5)}=\frac{2}{(x-2)(x+5)}$

\(\dfrac{x}{x-3}+\dfrac{-9}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{-9}{x\left(x-3\right)}=\dfrac{x^2-9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)}=\dfrac{x+3}{x}\)

\(\dfrac{x-5}{x^2-4x+4}:\dfrac{x^2-25}{2x-4}=\dfrac{x-5}{\left(x-2\right)^2}.\dfrac{2\left(x-2\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{2}{\left(x-2\right)\left(x+5\right)}\)

\(\dfrac{x+5}{x^2-5x}-\dfrac{x-5}{2x^2+10x}=\dfrac{x+25}{2x^2-50}\)

 \(\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x-5}{2x\left(x+5\right)}=\dfrac{x+5}{2\left(x-5\right)\left(x+5\right)}\)

dkxd : x ≠ 0

          x ≠ 5

          x ≠ -5

MTC : 2x(x - 5)(x + 5)

Quy đồng mẫu thức hai vế của phương trình :

⇒ \(\dfrac{2\left(x-5\right)\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}\) = \(\dfrac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}\)

Suy ra : 2(x - 5)(x + 5) - (x - 5)(x + 5) = x(x + 25)

         \(\Leftrightarrow\) 2(x² - 25) - (x² - 25) = x² + 25x

         \(\Leftrightarrow\) 2x² - 50 - x² + 25 - x² - 25x = 0

        \(\Leftrightarrow\) -25 - 25x = 0

        \(\Leftrightarrow\) -25x = 25

        \(\Leftrightarrow\) x = \(\dfrac{25}{-25}=-1\) (thỏa mãn)

 Vậy S = \(\left\{-1\right\}\)

 Chúc bạn học tốt

Ta có: \(\dfrac{x+5}{x^2-5x}-\dfrac{x-5}{2x^2+10x}=\dfrac{x+25}{2x^2-50}\)

\(\Leftrightarrow\dfrac{2\left(x+5\right)^2}{2x\left(x+5\right)\left(x-5\right)}-\dfrac{\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}=\dfrac{x\left(x+25\right)}{2x\left(x+5\right)\left(x-5\right)}\)

Suy ra: \(2\left(x^2+10x+25\right)-\left(x^2-10x+25\right)=x^2+25x\)

\(\Leftrightarrow2x^2+20x+50-x^2+10x-25-x^2-25x=0\)

\(\Leftrightarrow15x+25=0\)

\(\Leftrightarrow15x=-25\)

hay \(x=-\dfrac{5}{3}\)(thỏa ĐK)

\(x\ne0;x\ne\pm5\)

PT \(\Leftrightarrow\dfrac{x+25}{2\left(x+5\right)\left(x-5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}=0\)

\(\Rightarrow x^2+25x-2x^2-20x-50+x^2-10x+25=0\)

\(\Leftrightarrow-5x-25=0\)

\(\Leftrightarrow x=-5\) (ktm)
Vậy pt vô nghiệm.

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\x\ne\pm5\end{matrix}\right.\).

\(PT\Leftrightarrow\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}=\dfrac{5-x}{2x\left(x+5\right)}\)

\(\Leftrightarrow\dfrac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}-\dfrac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{\left(5-x\right)\left(x-5\right)}{2x\left(x-5\right)\left(x+5\right)}\)

\(\Rightarrow x\left(x+25\right)-2\left(x+5\right)^2=\left(5-x\right)\left(x-5\right)\)

\(\Leftrightarrow x^2+25x-2\left(x^2+10x+25\right)=10x-x^2-25\)

\(\Leftrightarrow-5x=25\Leftrightarrow x=-5\) (loại)

Vậy PT vô nghiệm