Đáp án:

 16) \( - \dfrac{{a\left( {a - b} \right)}}{{a + b}}\)

Giải thích các bước giải:

\(\begin{array}{l}
10)K = \dfrac{{x + x + 2}}{{x + 2}}:\dfrac{{4 - {x^2} - 3{x^2}}}{{\left( {2 - x} \right)\left( {x + 2} \right)}}\\
 = \dfrac{{2x + 2}}{{x + 2}}.\dfrac{{\left( {2 - x} \right)\left( {x + 2} \right)}}{{4\left( {1 - {x^2}} \right)}}\\
 = \dfrac{{2\left( {x + 1} \right)}}{{x + 2}}.\dfrac{{\left( {2 - x} \right)\left( {x + 2} \right)}}{{4\left( {x + 1} \right)\left( {1 - x} \right)}}\\
 = \dfrac{{2 - x}}{{2\left( {1 - x} \right)}}\\
11)M = \dfrac{{4{a^2} - b}}{{2{b^2}}}.\dfrac{{2a + b - 2a + b}}{{\left( {2a - b} \right)\left( {2a + b} \right)}}\\
 = \dfrac{{\left( {2a - b} \right)\left( {2a + b} \right)}}{{2{b^2}}}.\dfrac{{2b}}{{\left( {2a - b} \right)\left( {2a + b} \right)}}\\
 = \dfrac{1}{b}\\
12)N = \dfrac{{x - 1 - x}}{{x - 1}}:\dfrac{{{x^2} - 2x + 1 - 2x + 3}}{{{{\left( {x - 1} \right)}^2}}}\\
 = \dfrac{{ - 1}}{{x - 1}}.\dfrac{{{{\left( {x - 1} \right)}^2}}}{{{x^2} - 4x + 4}}\\
 = \dfrac{{ - 1}}{{x - 1}}.\dfrac{{{{\left( {x - 1} \right)}^2}}}{{{{\left( {x - 2} \right)}^2}}} =  - \dfrac{{x - 1}}{{{{\left( {x - 2} \right)}^2}}}\\
13)P = \dfrac{{4 + x\left( {x - 2} \right)}}{{x\left( {x - 2} \right)\left( {x + 2} \right)}}:\dfrac{{2\left( {x - 2} \right) - {x^2}}}{{2x\left( {x + 2} \right)}}\\
 = \dfrac{{{x^2} - 2x + 4}}{{x\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{2x\left( {x + 2} \right)}}{{ - {x^2} + 2x - 4}}\\
 =  - \dfrac{{{x^2} - 2x + 4}}{{x\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{2x\left( {x + 2} \right)}}{{{x^2} - 2x + 4}}\\
 =  - \dfrac{2}{{x - 2}}\\
14)Q = \dfrac{{x + 1 - x + 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{{{\left( {x + 1} \right)}^2}}}{4}\\
 = \dfrac{2}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{{{\left( {x + 1} \right)}^2}}}{4}\\
 = \dfrac{{x + 1}}{{2\left( {x - 1} \right)}}\\
15)R = \dfrac{8}{{{x^3} + 1}} + \dfrac{{4x\left( {1 - {x^2}} \right)}}{{{x^3} + 1}}.\dfrac{{x + 1 - x + 1}}{{\left( {{x^2} - 1} \right)\left( {x - 1} \right)}}\\
 = \dfrac{8}{{{x^3} + 1}} - \dfrac{{4x.2}}{{{x^3} + 1}}\\
 = \dfrac{{8 - 8x}}{{{x^3} + 1}}\\
16)S = \dfrac{{{a^2}\left( {a + b} \right) - {a^3}}}{{{{\left( {a + b} \right)}^2}}}:\dfrac{{a\left( {a - b} \right) - {a^2}}}{{\left( {a - b} \right)\left( {a + b} \right)}}\\
 = \dfrac{{{a^2}b}}{{{{\left( {a + b} \right)}^2}}}.\dfrac{{\left( {a - b} \right)\left( {a + b} \right)}}{{ - ab}}\\
 =  - \dfrac{{a\left( {a - b} \right)}}{{a + b}}
\end{array}\)