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 ...

ĐKXĐ: \(x\ne\left\{-3;-2;-1;0\right\}\)

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

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

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

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

\(\Leftrightarrow x=3\)

\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10

=>8(x+1/x)^2+4[(x+1/x)^2-2]^2-4[(x+1/x)^2-2](x+1/x)^2=(x+4)^2

Đặt x+1/x=a(a>=2)

=>8a^2+4[a^2-2]^2-4[a^2-2]*a^2=(x+4)^2

=>8a^2+4a^4-16a^2+16-4a^4+8a^2=(x+4)^2

=>(x+4)^2=16

=>x+4=4 hoặc x+4=-4

=>x=-8;x=0

Điều kiện: \(x\ne0\)

\(\Leftrightarrow8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left[\left(x^2+\dfrac{1}{x^2}\right)-\left(x+\dfrac{1}{x}\right)^2\right]=\left(x+4\right)^2\)

\(\Leftrightarrow8\left(x+\dfrac{1}{x}\right)^2-8\left(x^2+\dfrac{1}{x^2}\right)=\left(x+4\right)^2\\ \Leftrightarrow\left(x+4\right)^2=16\\ \Rightarrow\left\{{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)

Vì \(x\ne0\) nên \(S=\left\{-8\right\}\)

viết p/s như nào vậy bạn

`1/(3-x)-1/(x+1)=x/(x-3)-(x-1)^2/(x^2-2x-3)(x ne -1,3)`

`<=>(-x-1)/(x^2-2x-3)-(x-3)/(x^2-2x-3)=(x^2+x)/(x^2-2x-3)-(x-1)^2/(x^2-2x-3)`

`<=>-x-1-x+3=x^2+x-x^2+2x-1`

`<=>-2x+2=3x-1`

`<=>5x=3`

`<=>x=3/5`

Vậy `S={3/5}`

`1/(x-2)-6/(x+3)=6/(6-x^2-x)(x ne 2,-3)`

`<=>(x+3)/(x^2+x-6)-(6x-12)/(x^2+x-6)+6/(x^2+x-6)=0`

`<=>x+3-6x+12+6=0`

`<=>-5x+21=0`

`<=>x=21/5`

Vậy `S={21/5}`

a) ĐKXĐ: \(x\notin\left\{3;-1\right\}\)

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

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

Suy ra: \(-x-1-x+3=x^2+x-x^2+2x-1\)

\(\Leftrightarrow3x-1=-2x+2\)

\(\Leftrightarrow3x+2x=2+1\)

\(\Leftrightarrow5x=3\)

hay \(x=\dfrac{3}{5}\)(nhận)

Vậy: \(S=\left\{\dfrac{3}{5}\right\}\)