`1)` Biểu thức xác định `<=>x+1 \ne 0<=>x \ne -1`

`[x^2+2x+1]/[x+1]=[(x+1)^2]/[x+1]=x+1`

`2)` Bth xác định `<=>x(x-3) \ne 0<=>{(x \ne 0),(x \ne 3):}`

`[x^2-6x+9]/[x(x-3)]=[(x-3)^]/[x(x-3)]=[x-3]/x`

`3)` Bth xác định `<=>2x(x+2) \ne 0<=>{(x \ne 0),(x \ne -2):}`

`[x^2-4]/[2x(x+2)]=[(x-2)(x+2)]/[2x(x+2)]=[x-2]/[2x]`

`4)` Bth xác định `<=>5x^2-10x \ne 0<=>5x(x-2) \ne 0<=>{(x \ne 0),(x \ne 2):}`

`[x^2-2x]/[5x^2-10x]=[x(x-2)]/[5x(x-2)]=1/5`

1)

\(ĐKXĐ:x\ne-1\)

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

2)

ĐKXĐ x khác 0 và x khác 3

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

3)

ĐKXĐ: x khác 0 và x khác -2

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

4)

DKXĐ: x khác 0 và x khác 2

\(\dfrac{x^2-2x}{5x^2-10x}\\ =\dfrac{x\left(x-2\right)}{5x\left(x-2\right)}\\ =\dfrac{1}{5}\)

đk `x≠-1`

`(x^2+2x+1)/(x+1)`

`=((x+1)^2)/(x+1)`

`=x+1`

---------

đk \(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne3\end{matrix}\right.\)

`(x^2-6x+9)/(x(x-3))`

`=((x-3)^2)/(x(x-3))`

`=(x-3)/x`

--------

 đk \(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-2\end{matrix}\right.\)

`(x^2-4)/(2x(x+2))`

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

`=(x-2)/(2x)`

--------

đk \(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne2\end{matrix}\right.\)

`(x^2-2x)/(5x^2-10x)`

`=(x(x-2))/(5x(x-2))`

`=x/(5x)`

Bạn chú ý đăng lẻ câu hỏi! 1/

a/ \(=x^3-2x^5\)

b/\(=5x^2+5-x^3-x\)

c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)

d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)

e/ \(=x^4-x^2+2x^3-2x\)

f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)

cmm

Bài 1

A= (x-2)(2x-1)-2x(x+3)=2x²-x-4x+2-2x²-6x=-11x+2

Bài 1:

a) \(A=\left(x-2\right)\left(2x-1\right)-2x\left(x+3\right)\)

\(A=2x^2-x-4x+2-2x^2-6x\)

\(A=-11x+2\)

b) \(B=\left(3x-2\right)\left(2x+1\right)-\left(6x-1\right)\left(x+2\right)\)

\(B=6x^2+3x-4x-2-6x^2-12x+x+2\)

\(B=-12x\)

c) \(C=6x\left(2x+3\right)-\left(4x-1\right)\left(3x-2\right)\)

\(C=12x^2+18x-12x^2+8x+3x-2\)

\(C=29x-2\)

d) \(D=\left(2x+3\right)\left(5x-2\right)+\left(x+4\right)\left(2x-1\right)-6x\left(2x-3\right)\)

\(D=10x^2-4x+15x-6+2x^2-x+8x-4-12x^2+18x\)

\(D=36x-10\)

Bài 2: 

a: Ta có: \(2x\left(3x-5\right)\left(x+11\right)-3x\left(2x+3\right)\left(x+7\right)\)

\(=2x\left(3x^2+33x-5x-55\right)-3x\left(2x^2+14x+3x+21\right)\)

\(=6x^3+56x^2-110x-6x^2-51x^2-63x\)

\(=-117x\)

b: Ta có: \(\left(x^2+5x-6\right)\left(x-1\right)-\left(x+2\right)\left(x^2-x+1\right)-x\left(3x-10\right)\)

\(=x^3+4x^2-11x+6-\left(x^3-x^2+x+2x^2-2x+2\right)-3x^2+10x\)

\(=x^3+x^2-x+6-x^3-x^2+x-2\)

=4

c: Ta có: \(\left(x^2+x+1\right)\left(x-1\right)-x^2\left(x+1\right)+x^2-5\)

\(=x^3-1-x^3-x^2+x^2-5\)

=-6

a: \(=\dfrac{3\left(x-2\right)}{\left(x-2\right)^3}=\dfrac{3}{\left(x-2\right)^2}\)

b: \(=\dfrac{x^2\left(x+2\right)}{\left(x+2\right)^3}=\dfrac{x^2}{\left(x+2\right)^2}\)

\(B=\left(2x-3\right)\left(4x^2+6x+9\right)-\left(x+1\right)^2-\left(x-2\right)^3\)

\(=8x^3-27-x^2-2x-1-x^3+6x^2-12x+8\)

\(=7x^3+5x^2-14x-20\)

(x + 2)(x - 2) - (x - 2)(x + 5)

= (x - 2)(x + 2 - x - 5)

= (x - 2)-3

= -3x + 6

b) 2x(3x²y + 4x²y - 3)

= 2x(7x²y - 3)

= 14x³y - 6x

bạn giải hết đc ko ạ