Giúp e với ạk ạk … Hứa sẽ vote+cảm Ơn+ctlhn ạk ($\frac{3x}{1-3x}$+$\frac{2x}{3x+1}$):$\frac{6x^{2}+10x}{1-6x+9x^{2}}$

Đáp án:

\(\dfrac{{3x – 1}}{{2\left( {3x + 1} \right)}}\)

Giải thích các bước giải:

\(\begin{array}{l}
\left( {\dfrac{{3x}}{{1 – 3x}} + \dfrac{{2x}}{{3x + 1}}} \right).\dfrac{{{{\left( {1 – 3x} \right)}^2}}}{{2x\left( {3x + 5} \right)}}\\
 = \left[ {\dfrac{{3x\left( {3x + 1} \right) – 2x\left( {3x – 1} \right)}}{{\left( {3x + 1} \right)\left( {3x – 1} \right)}}} \right].\dfrac{{{{\left( {1 – 3x} \right)}^2}}}{{2x\left( {3x + 5} \right)}}\\
 = \dfrac{{9{x^2} + 3x – 6{x^2} + 2x}}{{\left( {3x + 1} \right)\left( {3x – 1} \right)}}.\dfrac{{{{\left( {3x – 1} \right)}^2}}}{{2x\left( {3x + 5} \right)}}\\
 = \dfrac{{3{x^2} + 5x}}{{\left( {3x + 1} \right)\left( {3x – 1} \right)}}.\dfrac{{{{\left( {3x – 1} \right)}^2}}}{{2x\left( {3x + 5} \right)}}\\
 = \dfrac{{x\left( {3x + 5} \right)}}{{\left( {3x + 1} \right)\left( {3x – 1} \right)}}.\dfrac{{{{\left( {3x – 1} \right)}^2}}}{{2x\left( {3x + 5} \right)}}\\
 = \dfrac{{x\left( {3x – 1} \right)}}{{2x\left( {3x + 1} \right)}} = \dfrac{{3x – 1}}{{2\left( {3x + 1} \right)}}
\end{array}\)