Question

Answer 1

Bạn ơi mấy câu đấy dài lắm bạn chịu khó đăng từng câu 1 nhá

Answer 2

Mình mún làm lắm mà dài quá nhưng bệnh lười tái phát r

Answer 3

a, 

\(\dfrac{x+3}{x^2-4}.\)\(\dfrac{8-12x+6x^2-x^3}{9x+27}\)=\(\dfrac{x+3}{\left(x-2\right)\left(x+2\right)}\).\(\dfrac{-\left(x-2\right)^3}{9\left(x+3\right)}\)

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

 

b,

\(\dfrac{6x-3}{5x^2+x}\).\(\dfrac{25x^2+10x+1}{1-8x^3}\)\(=\dfrac{3\left(2x-1\right)}{x\left(5x+1\right)}.\)\(\dfrac{\left(5x+1\right)^2}{1-\left(2x\right)^3}\)

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

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

 

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

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

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

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

 

d, \(\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}\)

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

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