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13 tháng 12 2020

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

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

  = \(\dfrac{-4+25-2x^2-x+2x^2+10x}{x^2-25}\)

  = \(\dfrac{21+9x}{x^2-25}\)

13 tháng 12 2020

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

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

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

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

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

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

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

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

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

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

 

17 tháng 1 2023

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

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

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

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

\(\Leftrightarrow-x=-44\)

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

\(2,\left(đk:x\ne4\right)\)

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

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

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

\(\Leftrightarrow-x=13\)

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

17 tháng 1 2023

bn ơi ktra lại câu 2 giúp mk đc ko 

AH
Akai Haruma
Giáo viên
11 tháng 1 2022

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

* Coi đây là bài toán rút gọn

Lời giải:

ĐKXĐ: $x\neq 2$

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

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

 

NV
3 tháng 1 2022

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

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

19 tháng 7 2021

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

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

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)

26 tháng 5 2021

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

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.

26 tháng 5 2021

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

 

 

26 tháng 11 2017

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

\(b,\dfrac{3x+5}{x^2-5x}+\dfrac{x-25}{5x-25}\)

\(=\dfrac{3x+5}{x\left(x-5\right)}+\dfrac{x-25}{5\left(x-5\right)}\)

\(=\dfrac{5\left(3x+5\right)}{5x\left(x-5\right)}+\dfrac{\left(x-25\right)x}{5x\left(x-5\right)}\)

\(=\dfrac{15x+25+x^2-25x}{5x\left(x-5\right)}\)

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

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

\(c,\left(\dfrac{2}{x-1}-\dfrac{2}{x+1}\right).\dfrac{x^2+2x+1}{4}\)

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

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

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

\(=\dfrac{x+1}{x-1}\)