\(\frac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}=\frac{\left(x^2\right)^2+2.x^2.3x+\left(3x\right)^2-1}{\left(x^2\right)^2+2.x^2.3x+\left(3x\right)^2-2x^2-6x+1}\)

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

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

Tao không biết

Lưu lê thanh hạ rảnh lên à bạn ???

\(\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)

\(=\left(6x+1+6x-1\right)^2\)

\(=36x^2\)

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

\(3x\left(x-3\right)-5\left(x-x^2+x^2-3\right)\)

\(3x\left(x-3\right)-5\left(x-3\right)\)

\(\left(x-3\right)\left(3x-5\right)\)

câu a đề có sai số mũ ko vậy

b) \(\dfrac{x^4+x^3-x-1}{x^4+x^3+2x^2+x+1}\)

\(=\dfrac{x^3\left(x+1\right)-\left(x+1\right)}{x^4+x^3+x^2+x^2+x+1}\)

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

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

c) \(\dfrac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}\)

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

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

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

a)

\(\dfrac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}\)

\(=\dfrac{x^2\left(x+1\right)-4\left(x+1\right)}{x^3+2x^2+6x^2+12x+5x+10}\)

\(=\dfrac{\left(x+1\right)\left(x^2-4\right)}{x^2\left(x+2\right)+6x\left(x+2\right)+5\left(x+2\right)}\)

\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x^2+6x+5\right)}\)

\(=\dfrac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left[x\left(x+5\right)+\left(x+5\right)\right]}\)

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

\(=\dfrac{x-2}{x+5}\)

b)

\(\dfrac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}\)

\(=\dfrac{x^4+3x^3+x^2+3x^3+9x^2+3x-x^2-3x-1}{x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1}\)

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

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

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

`(6x+1)^2-2(1+6x)(6x-1)+(6x-1)^2`

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

`=2^2=4`