Đáp án:
\(\begin{array}{l}
13)A13 = \dfrac{{x - 29}}{{{{\left( {x - 1} \right)}^2}}}\\
14)A14 = \dfrac{{ - 2{x^3} - 2{x^2} + 3x - 2}}{{2\left( {1 - {x^2}} \right)\left( {x - 2} \right)}}\\
15)A15 = \dfrac{{x - 1}}{2}\\
16)A16 = \dfrac{{x - 4}}{{x - 2}}\\
17)A17 = \dfrac{1}{{x + 5}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
13)DK:x \ne \pm 1\\
A13 = \dfrac{{3x - 31}}{{{{\left( {x - 1} \right)}^2}}} + \dfrac{{2\left( {x + 1} \right)}}{{\left( {1 - x} \right)\left( {x + 1} \right)}}\\
= \dfrac{{3x - 31}}{{{{\left( {x - 1} \right)}^2}}} - \dfrac{2}{{x - 1}}\\
= \dfrac{{3x - 31 - 2\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}\\
= \dfrac{{x - 29}}{{{{\left( {x - 1} \right)}^2}}}\\
14)DK:x \ne \left\{ { - 1;1;2} \right\}\\
A14 = \dfrac{x}{{x - 2}} + \dfrac{{{x^2} + 1}}{{2\left( {1 - {x^2}} \right)}}\\
= \dfrac{{x\left( {2 - 2{x^2}} \right) + \left( {{x^2} + 1} \right)\left( {x - 2} \right)}}{{2\left( {1 - {x^2}} \right)\left( {x - 2} \right)}}\\
= \dfrac{{2x - 2{x^3} + {x^3} - 2{x^2} + x - 2}}{{2\left( {1 - {x^2}} \right)\left( {x - 2} \right)}}\\
= \dfrac{{ - 2{x^3} - 2{x^2} + 3x - 2}}{{2\left( {1 - {x^2}} \right)\left( {x - 2} \right)}}\\
15)DK:x \ne \left\{ { - 5;0} \right\}\\
A15 = \dfrac{{x\left( {{x^2} + 2x} \right) + 2\left( {x + 5} \right)\left( {x - 5} \right) + 50 - 5x}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{{x^3} + 2{x^2} + 2{x^2} - 50 + 50 - 5x}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{{x^3} + 4{x^2} - 5x}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{{x^2} + 4x - 5}}{{2\left( {x + 5} \right)}}\\
= \dfrac{{\left( {x + 5} \right)\left( {x - 1} \right)}}{{2\left( {x + 5} \right)}} = \dfrac{{x - 1}}{2}\\
16)DK:x \ne \left\{ { - 3;2} \right\}\\
A16 = \dfrac{{\left( {x + 2} \right)\left( {x - 2} \right) - 5 - x - 3}}{{\left( {x - 2} \right)\left( {x + 3} \right)}}\\
= \dfrac{{{x^2} - 4 - x - 8}}{{\left( {x - 2} \right)\left( {x + 3} \right)}}\\
= \dfrac{{\left( {x - 4} \right)\left( {x + 3} \right)}}{{\left( {x - 2} \right)\left( {x + 3} \right)}}\\
= \dfrac{{x - 4}}{{x - 2}}\\
17)DK:x \ne \pm 5\\
A17 = \dfrac{{x - 5 + 2\left( {x + 5} \right) - 2x - 10}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}\\
= \dfrac{{ - x - 15 + 2x + 10}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}\\
= \dfrac{{x - 5}}{{\left( {x - 5} \right)\left( {x + 5} \right)}} = \dfrac{1}{{x + 5}}
\end{array}\)
