\(ĐKXĐ:1-3x\ne0\Leftrightarrow x\ne\dfrac{1}{3};3x+1\ne0\Leftrightarrow x\ne-\dfrac{1}{3}\)
\(A=\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x}\)
\(A=\left(\dfrac{9x^2+3x+2x-6x^2}{\left(1-3x\right)\left(3x+1\right)}\right):\left(\dfrac{6x^2+10x}{1+3x}\right)\)
\(A=\dfrac{3x^2+5x}{3x+1-9x^2-3x}.\dfrac{1+3x}{6x^2+10}\)
\(A=\dfrac{\left(3x^2+5x\right).\left(1+3x\right)}{\left(1-9x^2\right).2.\left(3x^2+5\right)}\)
\(A=\dfrac{1+3x}{\left(1+3x\right)\left(1-3x\right)}=\dfrac{1}{\left(1-3x\right).2}=\dfrac{1}{2-6x}\)
Hình như đề bị thiếu mũ \(2\) trong \(1-6x+9x\)\(\) đúng không Đinh Diệp
\(A=\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)
\(A=\left(\dfrac{3x\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{2x\left(1-3x\right)}{\left(1+3x\right)\left(1-3x\right)}\right):\dfrac{2x\left(3x+5\right)}{\left(3x-1\right)^2}\)
\(A=\left[\dfrac{3x\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{2x\left(1-3x\right)}{\left(1+3x\right)\left(1-3x\right)}\right]:\dfrac{2x\left(3x+5\right)}{\left(3x-1\right)^2}\)
\(A=\left[\dfrac{3x+9x^2}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{2x-6x^2}{\left(1+3x\right)\left(1-3x\right)}\right]:\dfrac{2x\left(3x+5\right)}{\left(3x-1\right)^2}\)
\(A=\left[\dfrac{3x+9x^2+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}\right].\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)
\(A=\left[\dfrac{3x^2+5x}{-\left(3x-1\right)\left(1+3x\right)}\right].\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)
\(A=\left[\dfrac{x\left(3x+5\right)}{-\left(3x-1\right)\left(1+3x\right)}\right].\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)
\(A=\dfrac{1}{-\left(1+3x\right)}.\dfrac{3x-1}{x}\)
\(A=\dfrac{1}{-1-3x}.\dfrac{3x-1}{x}\)
\(A=\dfrac{3x-1}{-x-3x^2}\)
