Đáp án:
\(\begin{array}{l}
a)DK:x \ne \left\{ { - 5;0} \right\}\\
b)\dfrac{{x - 1}}{2}\\
c)P = 0\\
\to x = 1\\
P = \dfrac{1}{4}\\
\to x = \dfrac{3}{2}\\
d)P > 0\\
\to x > 1\\
P < 0\\
\to x < 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ { - 5;0} \right\}\\
b)P = \dfrac{{{x^2} + 2x}}{{2x + 10}} + \dfrac{{x - 5}}{x} + \dfrac{{50 - 5x}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{x\left( {{x^2} + 2x} \right) + 2\left( {x + 5} \right)\left( {x - 5} \right) + 50 - 5x}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{{x^3} + 2{x^2} + 2{x^2} - 50 + 50 - 5x}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{{x^3} + 4{x^2} - 5x}}{{2x\left( {x + 5} \right)}}\\
= \dfrac{{{x^2} + 4x - 5}}{{2\left( {x + 5} \right)}}\\
= \dfrac{{\left( {x + 5} \right)\left( {x - 1} \right)}}{{2\left( {x + 5} \right)}}\\
= \dfrac{{x - 1}}{2}\\
c)P = 0\\
\to \dfrac{{x - 1}}{2} = 0\\
\to x - 1 = 0\\
\to x = 1\\
P = \dfrac{1}{4}\\
\to \dfrac{{x - 1}}{2} = \dfrac{1}{4}\\
\to 2\left( {x - 1} \right) = 1\\
\to x - 1 = \dfrac{1}{2}\\
\to x = \dfrac{3}{2}\\
d)P > 0\\
\to \dfrac{{x - 1}}{2} > 0\\
\to x - 1 > 0\\
\to x > 1\\
P < 0\\
\to \dfrac{{x - 1}}{2} < 0\\
\to x - 1 < 0\\
\to x < 1
\end{array}\)
