Đáp án:
d) \(\dfrac{{{{\left( {x + 3} \right)}^2}}}{{{{\left( {x + 2} \right)}^2}}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\left[ {\dfrac{1}{{x\left( {x + 1} \right)}} - \dfrac{{2 - x}}{{x + 1}}} \right]:\left( {\dfrac{{1 + {x^2} - 2x}}{x}} \right)\\
= \dfrac{{1 - x\left( {2 - x} \right)}}{{x\left( {x + 1} \right)}}.\dfrac{x}{{{{\left( {x - 1} \right)}^2}}}\\
= \dfrac{{1 - 2x + {x^2}}}{{x\left( {x + 1} \right)}}.\dfrac{x}{{{{\left( {x - 1} \right)}^2}}}\\
= \dfrac{{{{\left( {x - 1} \right)}^2}}}{{x\left( {x + 1} \right)}}.\dfrac{x}{{{{\left( {x - 1} \right)}^2}}}\\
= \dfrac{1}{{x + 1}}\\
c)\left[ {\dfrac{{9 + x\left( {x - 3} \right)}}{{x\left( {x + 3} \right)\left( {x - 3} \right)}}} \right]:\left[ {\dfrac{{x - 3}}{{x\left( {x + 3} \right)}} - \dfrac{x}{{3\left( {x + 3} \right)}}} \right]\\
= \dfrac{{{x^2} - 3x + 9}}{{x\left( {x + 3} \right)\left( {x - 3} \right)}}:\left[ {\dfrac{{3x - 9 - {x^2}}}{{3x\left( {x + 3} \right)}}} \right]\\
= \dfrac{{{x^2} - 3x + 9}}{{x\left( {x + 3} \right)\left( {x - 3} \right)}}.\dfrac{{3x\left( {x + 3} \right)}}{{ - \left( {{x^2} - 3x + 9} \right)}}\\
= - \dfrac{3}{{x - 3}}\\
b)\left[ {\dfrac{{3\left( {3x + 1} \right) + 2x\left( {1 - 3x} \right)}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}}} \right].\dfrac{{{{\left( {1 - 3x} \right)}^2}}}{{2x\left( {3x + 5} \right)}}\\
= \dfrac{{9x + 3 + 2x - 6{x^2}}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}}.\dfrac{{{{\left( {1 - 3x} \right)}^2}}}{{2x\left( {3x + 5} \right)}}\\
= \dfrac{{3 + 11x - 6{x^2}}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}}.\dfrac{{{{\left( {1 - 3x} \right)}^2}}}{{2x\left( {3x + 5} \right)}}\\
= \dfrac{{\left( {3 + 11x - 6{x^2}} \right)\left( {1 - 3x} \right)}}{{2x\left( {3x + 5} \right)\left( {1 + 3x} \right)}}\\
d)\dfrac{{x + 1}}{{x + 2}}:\left[ {\dfrac{{x + 2}}{{x + 3}}.\dfrac{{x + 1}}{{x + 3}}} \right]\\
= \dfrac{{x + 1}}{{x + 2}}.\dfrac{{{{\left( {x + 3} \right)}^2}}}{{\left( {x + 1} \right)\left( {x + 2} \right)}}\\
= \dfrac{{{{\left( {x + 3} \right)}^2}}}{{{{\left( {x + 2} \right)}^2}}}
\end{array}\)
