Giúp e với ạk ạk … Hứa sẽ vote+cảm Ơn+ctlhn ạk
($\frac{3x}{1-3x}$+$\frac{2x}{3x+1}$):$\frac{6x^{2}+10x}{1-6x+9x^{2}}$
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:
\(\begin{array}{l}
\left( {\dfrac{{3x}}{{1 – 3x}} + \dfrac{{2x}}{{3x + 1}}} \right).\dfrac{{{{\left( {1 – 3x} \right)}^2}}}{{2x\left( {3x + 5} \right)}}\\
= \left[ {\dfrac{{3x\left( {3x + 1} \right) – 2x\left( {3x – 1} \right)}}{{\left( {3x + 1} \right)\left( {3x – 1} \right)}}} \right].\dfrac{{{{\left( {1 – 3x} \right)}^2}}}{{2x\left( {3x + 5} \right)}}\\
= \dfrac{{9{x^2} + 3x – 6{x^2} + 2x}}{{\left( {3x + 1} \right)\left( {3x – 1} \right)}}.\dfrac{{{{\left( {3x – 1} \right)}^2}}}{{2x\left( {3x + 5} \right)}}\\
= \dfrac{{3{x^2} + 5x}}{{\left( {3x + 1} \right)\left( {3x – 1} \right)}}.\dfrac{{{{\left( {3x – 1} \right)}^2}}}{{2x\left( {3x + 5} \right)}}\\
= \dfrac{{x\left( {3x + 5} \right)}}{{\left( {3x + 1} \right)\left( {3x – 1} \right)}}.\dfrac{{{{\left( {3x – 1} \right)}^2}}}{{2x\left( {3x + 5} \right)}}\\
= \dfrac{{x\left( {3x – 1} \right)}}{{2x\left( {3x + 1} \right)}} = \dfrac{{3x – 1}}{{2\left( {3x + 1} \right)}}
\end{array}\)
Đáp án:
$(\dfrac{3x}{1-3x}+ \dfrac{2x}{3x+1}):(\dfrac{6x²+10x}{1-6x+9x²}\\=[\dfrac{3x(3x+1)+2x(1-3x)}{(1-3x)(1+3x)}]:[\dfrac{2x(2x+5)}{(1-3x)²}]\\=[\dfrac{9x²+3x+2x-6x²}{(1-3x)(1+3x)}].[\dfrac{(1-3x)²}{2x(3x+5)}]\\=[\dfrac{3x²+5x}{(1-3x)(1+3x)}].[\dfrac{(1-3x)²}{2x(3x+5)}]\\=[\dfrac{x(3x+5)}{(1-3x)(1+3x)}].[\dfrac{(1-3x)²}{2x(3x+5)}]\\=\dfrac{1}{ 1+3x}.\dfrac{1-3x}{2}\\=\dfrac{1-3x}{2(1+3x)}\\= \dfrac{1-3x}{2+3x²}$
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