Đáp án:
C14:
c. x=-25
Giải thích các bước giải:
\(\begin{array}{l}
C13:\\
a.DK:x \ne \pm 3\\
b.P = \dfrac{{3x - 9 + x + 3 + 18}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{4x + 12}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{4}{{x - 3}}\\
c.P = 4\\
\to \dfrac{4}{{x - 3}} = 4\\
\to x - 3 = 1\\
\to x = 4\\
C14:\\
a.DK:x \ne \left\{ { - 5;0} \right\}\\
b.P = \dfrac{{{x^2}.x + \left( {2x - 10} \right).\left( {5x + 25} \right) + 5\left( {50 + 5x} \right)}}{{5x\left( {x + 5} \right)}}\\
= \dfrac{{{x^3} + 10{x^2} + 50x - 50x - 250 + 250 + 25x}}{{5x\left( {x + 5} \right)}}\\
= \dfrac{{{x^3} + 10{x^2} + 25x}}{{5x\left( {x + 5} \right)}}\\
= \dfrac{{x{{\left( {x + 5} \right)}^2}}}{{5x\left( {x + 5} \right)}}\\
= \dfrac{{x + 5}}{5}\\
c.P = - 4\\
\to \dfrac{{x + 5}}{5} = - 4\\
\to x + 5 = - 20\\
\to x = - 25
\end{array}\)
