Question

cứu với

Answer 1

1: ĐKXĐ: x∉{2;-2;3/4}

2: \(D=\left(\frac{x+2}{2x-4}+\frac{x-2}{2x+4}-\frac{8}{x^2-4}\right)\cdot\frac{3-4x}{x^2+1}\)

\(=\frac{\left(x+2\right)^2+\left(x-2\right)^2-16}{2\left(x-2\right)\left(x+2\right)}\cdot\frac{3-4x}{x^2+1}\)

\(=\frac{x^2+4x+4+x^2-4x+4-16}{2\left(x-2\right)\left(x+2\right)}\cdot\frac{3-4x}{x^2+1}=\frac{2x^2-8}{2\left(x^2-4\right)}\cdot\frac{3-4x}{x^2+1}=\frac{3-4x}{x^2+1}\)

\(\frac{1}{D}=\frac{x^2+1}{3-4x}\)

=>\(\frac{1}{D}+1=\frac{x^2+1+3-4x}{3-4x}=\frac{\left(x-2\right)^2}{3-4x}\ge0\forall x\) thỏa mãn ĐKXĐ

=>\(\frac{1}{D}\ge-1\forall x\) thỏa mãn ĐKXĐ

=>D<=-1∀x thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x-2=0

=>x=2(nhận)