Question

\(\frac{12x}{x^2+4x+2}\) + \(\frac{-3x}{x^2+2x+2}\)= 1

 

 

Answer 1

Đặt \(A=\frac{12x}{x^2+4x+2}-\frac{3x}{x^2+2x+2}=1\)

 \(\Rightarrow A=\frac{12x\left(x^2+2x+2\right)-3x\left(x^2+4x+2\right)}{\left(x^2+4x+2\right)\left(x^2+2x+2\right)}\)

   \(\Rightarrow A=\frac{12x^3+24x^2+24x-3x^3-12x^2-6x}{\left(x^2+4x+2\right)\left(x^2+2x+2\right)}\)

   \(\Rightarrow A=\frac{9x^3+12x^2+18x}{\left(x^2+4x+2\right)\left(x^2+2x+2\right)}\)

   \(\Rightarrow A=\frac{9x^3+12x^2+18x}{\left(x^2+4x+2\right)\left(x^2+2x+2\right)}\)

Answer 2

tìm x á ạ