ĐKXĐ: \(x\ne\left\{-4;-5;-6;-7\right\}\)

\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Rightarrow\left(x+4\right)\left(x+7\right)=54\)

\(\Leftrightarrow x^2+11x-26=0\)

\(\Leftrightarrow x^2-2x+13x-26=0\)

\(\Leftrightarrow x\left(x-2\right)+13\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)

Ta có:\(x^3+9x^2+11x-21\)

\(=x^3-x^2+10x^2-10x+21x-21=x^2\left(x-1\right)+10x\left(x-1\right)+21\left(x-1\right)\)

\(=\left(x^2+10x+21\right)\left(x-1\right)=\left(x^2+3x+7x+21\right)\left(x-1\right)\)

\(=\left[x\left(x+3\right)+7\left(x+3\right)\right]\left(x-1\right)\)

\(=\left(x+3\right)\left(x+7\right)\left(x-1\right)\)

x^3+9x^2+11x-21=x^3-x^2+10x^2-10x+21x-21=(x^3-x^2)+(10x^2-10x)+(21x-21)

=x^2(x-1)+10x(x-1)+21(x-1)=(x-1)(x^2+10x+21)=(x-1)(x^2+3x+7x+21)=(x-1)[(x^2+3x)+(7x+21)]

=(x-1)(x+7)(x+3)

\(x^3+9x^2+11x-21\)

\(=x^3-x^2+10x^2-10x+21x-21\)

\(=\left(x^3-x^2\right)+\left(10x^2-10x\right)+\left(21x-21\right)\)

\(=x^2\left(x-1\right)+10x\left(x-1\right)+21\left(x-1\right)\)

\(=\left(x^2+10x+21\right)\left(x-1\right)\)

\(=\left(x^2+10x+25-4\right)\left(x-1\right)\)

\(=\left[\left(x-5\right)^2-4\right]\left(x-1\right)\)

\(=\left(x-7\right)\left(x-3\right)\left(x-1\right)\)

ko chắc phần sau lắm nhé

`@` `\text {Ans}`

`\downarrow`

`4x^3 - 4x^2 - 9x + 9`

`= (4x^3 - 4x^2) - (9x - 9)`

`= 4x^2(x - 1) - 9(x - 1)`

`= (4x^2 - 9)(x - 1)`

____

`x^3 + 6x^2 + 11x + 6`

`= x^3 + x^2 + 5x^2 + 5x + 6x + 6`

`= (x^3 + x^2) + (5x^2 + 5x) + (6x + 6)`

`= x^2*(x + 1) + 5x(x + 1) + 6(x + 1)`

`= (x^2 + 5x + 6)(x+1)`

____

`x^2y - x^3 - 9y + 9x`

`= (x^2y - 9y) - (x^3 - 9x)`

`= y(x^2 - 9) - x(x^2 - 9)`

`= (y - x)(x^2 - 9)`

b: =x^3+x^2+5x^2+5x+6x+6

=(x+1)(x^2+5x+6)

=(x+1)(x+2)(x+3)

c: =x^2(y-x)-9(y-x)

=(y-x)(x^2-9)

=(y-x)(x-3)(x+3)

a: =(4x^3-4x^2)-(9x-9)

=4x^2(x-1)-9(x-1)

=(x-1)(4x^2-9)

=(x-1)(2x-3)(2x+3)

\(x^3+x^2+9x-10x^2-10x+25x+25\)

\(=x^2\left(x+1\right)-10x\left(x+1\right)+25\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-10x+25\right)=\left(x+1\right)\left(x-5\right)^2\)

tao dep trai lam

1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)

\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)

\(=x^2+4x+3-\dfrac{12}{2x+3}\)

a) Tìm MTC:

\(\text{2x + 6 = 2(x + 3)}\)

\(\text{x2 – 9 = (x – 3)(x + 3)}\)

\(\text{MTC = 2(x – 3)(x + 3) = 2(x2 – 9)}\)

Nhân tử phụ:

\(\text{2(x – 3)(x + 3) : 2(x + 3) = x – 3}\)

\(\text{2(x – 3)(x + 3) : (x2 – 9) = 2}\)

Qui đồng:

b) Tìm MTC: