Bài 1: Phân thức đại số.

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

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

Với \(x\ne-1\), ta có:

\(3x-1=0\Rightarrow3x=1\) \(\Rightarrow x=\dfrac{1}{3}\left(tm\right)\)

Thay \(x=\dfrac{1}{3}\) vào (1), ta được:

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

\(=-\dfrac{8}{3}:\dfrac{4}{3}=-\dfrac{8}{3}\cdot\dfrac{3}{4}=-2\)

Vậy: ...

2a(a-b)-3b(a-b)=(2a-3b)(a-b)

2a(a-b)-3b(a-b)=(2a-3b)(a-b)

a: \(9x^3y^2+3x^2y^2\)

\(=3x^2y^2\cdot3x+3x^2y^2\cdot1\)

\(=3x^2y^2\left(3x+1\right)\)

b: \(x^2-2x+1-y^2\)

\(=\left(x^2-2x+1\right)-y^2\)

\(=\left(x-1\right)^2-y^2\)

\(=\left(x-1-y\right)\left(x-1+y\right)\)

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

a: \(4x^2-4x\)

\(=4x\cdot x-4x\cdot1\)

\(=4x\left(x-1\right)\)

b: \(x^2-2xy+y^2-4\)

\(=\left(x-y\right)^2-2^2\)

\(=\left(x-y-2\right)\left(x-y+2\right)\)

`4x^2-4x`

`= 4x(x-1)`

__

`x^2 -2xy +y^2-4`

`= (x^2-2xy+y^2)-2^2`

`=(x-y)^2 -2^2`

`=(x-y-2)(x-y+2)`

a,

\(4x^2-4x\\=4x(x-1)\)

b,

\(x^2-2xy+y^2-4\\=(x^2-2xy+y^2)-4\\=(x-y)^2-2^2\\=(x-y-2)(x-y+2)\)

\(\text{#}Toru\)

\(\left(x-2\right)^2-x\left(3+x\right)=25\)

\(\Leftrightarrow\left(x^2-4x+4\right)-\left(3x+x^2\right)=25\)

\(\Leftrightarrow x^2-4x+4-3x-x^2=25\)

\(\Leftrightarrow-7x+4=25\)

\(\Leftrightarrow-7x=25-4\)

\(\Leftrightarrow-7x=21\)

\(\Leftrightarrow x=\dfrac{21}{-7}\)

\(\Leftrightarrow x=-3\)

Vậy: .... 

\(\dfrac{x^3+1}{x-9}-\dfrac{3x-9}{x+3}-x\)

\(=\dfrac{\left(x^3+1\right)\left(x+3\right)-\left(3x-9\right)\left(x-9\right)-x\left(x-9\right)\left(x+3\right)}{\left(x-9\right)\left(x+3\right)}\)

\(=\dfrac{x^4+3x^3+x+3-3x^2+27x+9x-81-x\left(x^2-6x-27\right)}{\left(x-9\right)\left(x+3\right)}\)

\(=\dfrac{x^4+37x-78-x^3+6x^2+27x}{\left(x-9\right)\left(x+3\right)}\)

\(=\dfrac{x^4-x^3+6x^2+64x-78}{\left(x-9\right)\left(x+3\right)}\)

\(\dfrac{2x^2+4x+8}{x^3-8}\)

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

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

\(=\dfrac{2}{x-2}\)