Đáp án:
b) \(\left[ \begin{array}{l}
x > 2\\
x < - 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)y' = {x^2} - 4x\\
y' > 0 \to {x^2} - 4x > 0\\
\to x\left( {x - 4} \right) > 0\\
\to \left[ \begin{array}{l}
x > 4\\
x < 0
\end{array} \right.\\
b)y' = 1 - \dfrac{4}{{{x^2}}}\\
y' > 0 \to 1 - \dfrac{4}{{{x^2}}} > 0\\
\to \dfrac{{{x^2} - 4}}{{{x^2}}} > 0\\
\to {x^2} - 4 > 0\left( {x \ne 0} \right)\\
\to \left[ \begin{array}{l}
x > 2\\
x < - 2
\end{array} \right.\\
c)DK:1 > x > - 1\\
y' = - 1.\dfrac{1}{{2\sqrt {1 - x} }} + 1.\dfrac{1}{{2\sqrt {1 + x} }}\\
= \dfrac{{ - 1}}{{2\sqrt {1 - x} }} + \dfrac{1}{{2\sqrt {1 + x} }}\\
= \dfrac{{ - \sqrt {1 + x} + \sqrt {1 - x} }}{{2\sqrt {1 - {x^2}} }}\\
y' > 0\\
\to \dfrac{{ - \sqrt {1 + x} + \sqrt {1 - x} }}{{2\sqrt {1 - {x^2}} }} > 0\\
\to - \sqrt {1 + x} + \sqrt {1 - x} > 0\\
\to \sqrt {1 - x} > \sqrt {1 + x} \\
\to 1 - x > 1 + x\\
\to 0 > 2x\\
\to 0 > x\\
\to 0 > x > - 1
\end{array}\)
