a: \(\frac{11x-3}{3x^2-15x-42}\)

\(=\frac{11x-3}{3\left(x^2-5x-14\right)}=\frac{11x-3}{3\left(x-7\right)\left(x+2\right)}\)

\(=\frac{3\left(11x-3\right)\left(x+1\right)}{3\cdot3\cdot\left(x-7\right)\left(x+2\right)\left(x+1\right)}=\frac{3\left(11x-3\right)\left(x+1\right)}{9\left(x-7\right)\left(x+2\right)\left(x+1\right)}\)

\(\frac{8}{x^2-6x-7}=\frac{8}{x^2-7x+x-7}\)

\(=\frac{8}{\left(x-7\right)\left(x+1\right)}\)

\(=\frac{8\cdot9\cdot\left(x+2\right)}{9\left(x+2\right)\left(x-7\right)\left(x+1\right)}=\frac{72x+144}{9\left(x+2\right)\left(x-7\right)\left(x+1\right)}\)

\(\frac{13x}{9x-63}=\frac{13x}{9\left(x-7\right)}\)

\(=\frac{13x\left(x+2\right)\left(x+1\right)}{9\left(x-7\right)\left(x+2\right)\left(x+1\right)}\)

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

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

\(\frac{3x^2-6x}{x^2-2x+4}=\frac{3x\left(x-2\right)}{x^2-2x+4}=\frac{3x\left(x-2\right)\cdot x\left(x+2\right)}{x\left(x+2\right)\left(x^2-2x+4\right)}\)

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

\(\frac{10x^2+28x-8}{x^4+8x}=\frac{10x^2+28x-8}{x\left(x^3+8\right)}=\frac{10x^2+28x-8}{x\left(x+2\right)\left(x^2-2x+4\right)}\)

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:

\(1-\dfrac{1}{x+1}=\dfrac{14}{15}\)

\(\dfrac{x+1-1}{x+1}=\dfrac{14}{15}\)

\(\dfrac{x}{x+1}=\dfrac{14}{15}\)

\(15x=14x+14\)

\(x=14\)

`(x+3)(x^2-5x+8)=(x+3).x^2`

`<=>(x+3)(x^2-5x+8-x^2)=0`

`<=>(x+3)(8-5x)=0`

`<=>` \(\left[ \begin{array}{l}x+3=0\\8-5x=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=\dfrac85\\x=-3\end{array} \right.\) 

Vậy `S={-3,8/5}`

`(x+3)(x^2-5x+8)=(x+3).x^2`

`<=>(x+3)(x^2-5x+8-x^2)=0`

`<=>(x+3)(-5x+8)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\-5x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{8}{5}\end{matrix}\right.\)

Vậy `S={-3;8/5}`.

`|x-2|=3-x`

`@TH1:x-2 >= 0<=>x >= 2=>|x-2|=x-2`

    `=>x-2=3-x`

`<=>2x=5`

`<=>x=5/2` (t/m)

`@TH2:x-2 < 0<=>x < 2=>|x-2|=2-x`

    `=>2-x=3-x`

`<=>0x=1` (Vô lí)

Vậy `S={5/2}`

\(\left|x-2\right|=3-x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=3-x\\x-2=x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2-3+x=0\\x-2-x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\\left(x-x\right)+\left(-2+3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\1=0\left(vl\right)\end{matrix}\right.\)

\(=>x=\dfrac{5}{2}\)

\(A=2x^3+6x^2-3x+\dfrac{1}{2}=2\cdot\dfrac{1}{3}^3+6\cdot\dfrac{1}{3}^2-3\cdot\dfrac{1}{3}+\dfrac{1}{2}\)

=13/54

    3(x+3)-x(x+3)=0

     (x+3)(3-x)     =0

      x+3 =0 hoặc 3-x=0   =>x={-3;3}