Đáp án:
10) \(\dfrac{1}{{2\left( {x + 1} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
2)DK:x \ne 0\\
\dfrac{{\left( {x + 2} \right).5.4 + 3.4.\left( {x + 5} \right) - 3.5\left( {x + 8} \right)}}{{3.5.4.x}}\\
= \dfrac{{20x + 40 + 12x + 60 - 15x - 120}}{{60x}}\\
= \dfrac{{17x - 20}}{{60x}}\\
4)DK:a \ne b;b \ne 0;a \ne 0\\
\dfrac{{b - a}}{{ab\left( {a - b} \right)}} = - \dfrac{{a - b}}{{ab\left( {a - b} \right)}} = - \dfrac{1}{{ab}}\\
6)DK:x \ne \pm 2\\
\dfrac{1}{{x + 2}} + \dfrac{2}{{\left( {x + 2} \right)\left( {x - 2} \right)}}\\
= \dfrac{{x - 2 + 2}}{{\left( {x + 2} \right)\left( {x - 2} \right)}} = \dfrac{x}{{\left( {x + 2} \right)\left( {x - 2} \right)}}\\
8)DK:x \ne \pm 2\\
\dfrac{{4\left( {x - 2} \right) + 2\left( {x - 2} \right) - 5x + 6}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{4x - 8 + 2x - 4 - 5x + 6}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{x - 6}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
10)DK:x \ne \pm 1\\
\dfrac{x}{{2\left( {x - 1} \right)}} + \dfrac{{{x^2} + 1}}{{2\left( {1 - {x^2}} \right)}}\\
= \dfrac{{x\left( {x + 1} \right) - {x^2} - 1}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{{x^2} + x - {x^2} - 1}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{x - 1}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{1}{{2\left( {x + 1} \right)}}
\end{array}\)
