Question

Answer 1

a: ĐKXĐ: x∉{-2;0}

b: \(B=\frac{\left(x+2\right)^2}{x}\cdot\left(1-\frac{x^2}{x+2}\right)-\frac{x^2+6x+4}{x}\)

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

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

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

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

c: \(B=-x^2-2x-2\)

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

\(=-\left(x^2+2x+1+1\right)=-\left(x+1\right)^2-1\le-1\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x+1=0

=>x=-1