Question

Answer 1

\(a\))\(Ta\) \(có\):

    \(A=\dfrac{\left(x+2\right)^2}{x}\left(1-\dfrac{x^2}{x+2}\right)\)  với \(x\ne0,x\ne-2\)

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

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

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

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

Vậy...

 

Answer 2

a) Ta có: \(A=\dfrac{\left(x+2\right)^2}{x}\cdot\left(1-\dfrac{x^2}{x+2}\right)\)

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

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

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

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

c) Ta có: \(C=\dfrac{A}{B}\)

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

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

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