Question

cho \(A=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)

voi \(x\ne\pm2\)

a) rut gon A

b)tim x de \(A-\frac{1}{x}=-1\)

Answer 1

a) \(A=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)

\(A=\frac{4x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5x-6}{\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{4x^2-8x+2x+4-5x+6}{\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{4x^2-11x+10}{\left(x-2\right)\left(x+2\right)}\)

Answer 2

\(a,A=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)

\(=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{6-5x}{x^2-4}\)

\(=\frac{4x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{4x^2-8x+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{4x^2-11x+10}{\left(x-2\right)\left(x+2\right)}\)

Answer 3

b) Ta có :

\(\frac{4x^2-11x+10}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x}=-1\)

\(\frac{1}{x}=\frac{4x^2-11x+10}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\frac{1}{x}=\frac{4x^2-11x+10+x^2-4}{\left(x-2\right)\left(x+2\right)}\)

\(\frac{1}{x}=\frac{5x^2-11x+6}{\left(x-2\right)\left(x+2\right)}\)

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

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

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

\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=x\left(x-1\right)\left(5x-6\right)\)

\(\Leftrightarrow x=\frac{\left(x+2\right)\left(x-2\right)}{\left(x-1\right)\left(5x-6\right)}\)