ĐKXĐ: \(x\notin\left\{10;-10\right\}\)

Ta có: \(\dfrac{720}{x+10}+4=\dfrac{720}{x-10}\)

\(\Leftrightarrow\dfrac{720\left(x-10\right)}{\left(x+10\right)\left(x-10\right)}+\dfrac{4\left(x^2-100\right)}{\left(x+10\right)\left(x-10\right)}=\dfrac{720\left(x+10\right)}{\left(x+10\right)\left(x-10\right)}\)

Suy ra: \(720x-7200+4x^2-400-720x-7200=0\)

\(\Leftrightarrow4x^2=14800\)

\(\Leftrightarrow x^2=3700\)

hay \(x\in\left\{10\sqrt{37};-10\sqrt{37}\right\}\)

ĐKXĐ: \(x\ne\pm10\)

\(\Leftrightarrow\dfrac{180}{x-10}-\dfrac{180}{x+10}=1\)

\(\Leftrightarrow\dfrac{180\left(x+10-x+10\right)}{\left(x-10\right)\left(x+10\right)}=1\)

\(\Leftrightarrow\dfrac{3600}{x^2-100}=1\)

\(\Rightarrow x^2-100=3600\)

\(\Leftrightarrow x^2=3700\)

\(\Leftrightarrow x=\pm10\sqrt{37}\) (thỏa mãn)

bài 1:

\(\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}\)

<=>\(\left(\dfrac{x-10}{1994}-1\right)+\left(\dfrac{x-8}{1996}+-1\right)+\left(\dfrac{x-6}{1998}-1\right)=\left(\dfrac{x-2002}{2}-1\right)+\left(\dfrac{x-2000}{4}-1\right)+\left(\dfrac{x-1998}{6}-1\right)\)

<=>\(\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}\)

<=>\(\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}=0\)

<=>(x-2004)\(\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}\right)\)

vì 1/1994+1/1996+1/1998-1/2-1/4-1/6 khác 0

nên x-2004=0=>x=2004

vyaj.......

bài 2:

\(\dfrac{x-85}{15}+\dfrac{x-74}{13}+\dfrac{x-67}{11}+\dfrac{x-64}{9}=10\)

<=>\(\left(\dfrac{x-85}{15}-1\right)+\left(\dfrac{x-74}{13}-2\right)+\left(\dfrac{x-67}{11}-3\right)+\left(\dfrac{x-64}{9}-4\right)=0\)

<=>\(\dfrac{x-100}{15}+\dfrac{x-100}{13}+\dfrac{x-100}{11}+\dfrac{x-100}{9}=0\)

<=>\(\left(x-100\right)\left(\dfrac{1}{15}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{9}\right)=0\)

vì 1/15+1/13+1/11+1/9 khác 0

=>x-100=0<=>x=100

điều kiện xác định \(x\ne0\)

ta có : \(\dfrac{x+1}{x^2+2x+4}-\dfrac{x-2}{x^2-2x+4}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x^2-2x+4\right)-\left(x-2\right)\left(x^2+2x+4\right)}{\left(x^2+2x+4\right)\left(x^2-2x+4\right)}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)

\(\Leftrightarrow-x^2+2x+12=\dfrac{6}{x}\Leftrightarrow x\left(-x^2+2x+12\right)=6\)

\(\Leftrightarrow-x^3+2x^2+12x=6\Leftrightarrow-x^3+2x^2+12x-6=0\)

tới đây bn bấm máy tính nha

ĐKXĐ \(x\ne8;x\ne11;x\ne9;x\ne10\)

\(\dfrac{8}{x-8}+\dfrac{11}{x-11}=\dfrac{9}{x-9}+\dfrac{10}{x-10}\)

\(\Leftrightarrow\left(\dfrac{8}{x-8}+1\right)+\left(\dfrac{11}{x-11}+1\right)=\left(\dfrac{9}{x-9}+1\right)+\left(\dfrac{10}{x-10}+1\right)\)

\(\Leftrightarrow\dfrac{x}{x-8}+\dfrac{x}{x-11}=\dfrac{x}{x-9}+\dfrac{x}{x-10}\)

\(\Leftrightarrow\dfrac{x}{x-8}+\dfrac{x}{x-11}-\dfrac{x}{x-9}-\dfrac{x}{x-10}=0\)

\(\Leftrightarrow x\left(\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10}\right)=0\)

\(\Leftrightarrow x=0\) hoặc \(\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10}=0\)

1) x=0

2) \(\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10}=0\)

\(\Leftrightarrow\dfrac{x-11+x-8}{\left(x-8\right)\left(x-11\right)}-\dfrac{x-10+x-9}{\left(x-9\right)\left(x-10\right)}=0\)

\(\Leftrightarrow\dfrac{2x-19}{\left(x-8\right)\left(x-11\right)}=\dfrac{2x-19}{\left(x-9\right)\left(x-10\right)}\)

\(\Leftrightarrow\dfrac{2x-19}{x^2-19x+88}=\dfrac{2x-19}{x^2-19x+90}\)

do \(x^2-19x+88\ne x^2-19x+90\)

\(\Rightarrow2x-19=0\)

=> x=\(\dfrac{19}{2}\)

Vậy x=\(0\); x=\(\dfrac{19}{2}\)

Tik

a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow x^2-2x+12-8-x^2=0\)

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow-2x=-4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

\(\Leftrightarrow\left|2x+6\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-3}

ĐK: ` x \ne 10; x \ne 0`

`120/(x-10)-3/5=120/x`

`<=>120/(x-10)-120/x=3/5`

`<=>1/(x-10) - 1/x= 1/200`

`<=> (x-x+10)/(x(x-10)) = 1/200`

`<=> 10/(x(x-10))= 1/200`

`<=> x^2-10=2000`

`<=>` \(\left[{}\begin{matrix}x=50\\x=-40\end{matrix}\right.\)

Vậy `S={50;-40}`.

`120/(x-10)-3/5=120/x(x ne 0,x ne 10)`

`<=>40/(x-10)-1/5=40/x`

`<=>200x-x(x-10)=200(x-10)`

`<=>200x-200x+2000-x^2+10x=0`

`<=>x^2-10x-2000=0`

`Delta'=25+2000=2025`

`<=>x_1=50,x_2=-40`

Vậy `S={50,-40}`

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne10\\x\ne0\end{matrix}\right.\)

Ta có : \(\dfrac{120}{x-10}-\dfrac{3}{5}=\dfrac{120}{x}=\dfrac{600-3\left(x-10\right)}{5\left(x-10\right)}\)

\(\Leftrightarrow600\left(x-10\right)=600x-3x\left(x-10\right)\)

\(\Leftrightarrow600x-6000=600x-3x^2+30x\)

\(\Leftrightarrow3x^2-30x-6000=0\)

\(\Leftrightarrow\left(x-50\right)\left(x+40\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=50\\x=-40\end{matrix}\right.\) ( TM )

Vậy ...