Question

Thực hiện phép tính sau:

\(\frac{x+1}{x-3}-\frac{1-x}{x+3}-\frac{2x\left(1-x\right)}{9-x^2}\)

GIÚP MK NAH~ THANKS!!!

Answer 1

 = (x+1).(x+3)-(1-x).(x-3)+2x.(1-x)/(x-3).(x+3)

 = x^2+4x+3+x^2-4x+3+2x-2x^2/(x+3).(x-3)

 = 2x+6/(x+3).(x-3) = 2.(x+3)/(x+3).(x-3) = 2/x-3

k mk nha

Answer 2

\(\frac{x+1}{x-3}\)\(-\)\(\frac{1-x}{x+3}\)\(-\)\(\frac{2x\left(1-x\right)}{9-x^2}\)

\(=\)\(\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)\(-\)\(\frac{\left(1-x\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)\(-\)\(\frac{2x\left(1-x\right)}{9-x^2}\)

\(=\)\(\frac{x^2+4x+3}{x^2-9}\)\(-\)\(\frac{4x-x^2-3}{x^2-9}\)\(+\)\(\frac{2x-2x^2}{x^2-9}\)

\(=\)\(\frac{x^2+4x+3-4x+x^2+3+2x-2x^2}{x^2-9}\)\(=\)\(\frac{6+2x}{\left(x-3\right)\left(x+3\right)}\)\(=\)\(\frac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\)\(\frac{2}{x-3}\)