Đáp án:
B15:
a. \(\dfrac{{ - 6}}{{x - 2}}\)
Giải thích các bước giải:
\(\begin{array}{l}
B15:\\
a.DK:x \ne \pm 2\\
P = \left[ {\dfrac{{x - 2\left( {x + 2} \right) + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}} \right].\left( {x + 2} \right)\\
= \dfrac{{2x - 2x - 2 - 4}}{{x - 2}}\\
= \dfrac{{ - 6}}{{x - 2}}\\
b.Thay:x = - 2000\\
\to P = \dfrac{{ - 6}}{{ - 2002}} = \dfrac{3}{{1001}}\\
c.P = \dfrac{1}{{19}}\\
\to \dfrac{{ - 6}}{{x - 2}} = \dfrac{1}{{19}}\\
\to x - 2 = - 114\\
\to x = - 112\\
d.P < 0\\
\to \dfrac{{ - 6}}{{x - 2}} < 0\\
\to x - 2 < 0\\
\to x < 2;x \ne - 2\\
B16:\\
a.DK:x \ne \left\{ {0;3} \right\}\\
Q = \left[ {\dfrac{x}{{3\left( {x - 3} \right)}} + \dfrac{{2x - 3}}{{x\left( {3 - x} \right)}}} \right].\dfrac{{3{x^2} - 9x}}{{{x^2} - 6x + 9}}\\
= \left[ {\dfrac{{{x^2} - 3\left( {2x - 3} \right)}}{{3x\left( {x - 3} \right)}}} \right].\dfrac{{3x\left( {x - 3} \right)}}{{{{\left( {x - 3} \right)}^2}}}\\
= \dfrac{{{x^2} - 6x + 9}}{{3x\left( {x - 3} \right)}}.\dfrac{{3x}}{{x - 3}}\\
= \dfrac{{{{\left( {x - 3} \right)}^2}}}{{3x\left( {x - 3} \right)}}.\dfrac{{3x}}{{x - 3}}\\
= \dfrac{{x - 3}}{{3x}}.\dfrac{{3x}}{{x - 3}} = 1
\end{array}\)
