trên gg có kết quả kìa

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:\(=3x^3-9x^2-5x^2+15x-x+3\)

\(=\left(x-3\right)\left(3x^2-5x-1\right)\)

2: \(=x^4-3x^3+x^2-3x^3+9x^2-3x+x^2-3x+1\)

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

a: =x^4-3x^5+4x^8

b: =2x^3+2x^2+4x

c: =4x^2+8x-5

d: =2x+3x^2+7x^4

câu d

\(D=\dfrac{\left(1-x^2\right)}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{\left(1-x^2\right)\left(x^2-x-3\right)+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{x^2-x-3-x^4+x^3-3x^2+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x^4+x^3+x^2-15x}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x\left(x^3-x^2-x+15\right)}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-\left(x^3-x^2-x+15\right)}{\left(x+3\right)}\end{matrix}\right.\)