Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
a)y = {\rm{ }}f\left( x \right) = \sqrt {x - 3} \\
dkxd:x \ge 3\\
Lay:{\rm{ }}{x_1}\; \ne {\rm{ }}{x_2}\; \in {\rm{ }}R,{\rm{ }}ta{\rm{ }}co:\\
\frac{{f({x_1}) - f({x_2})}}{{{x_1} - {x_2}}} = \frac{{\sqrt {{x_1} - 3} - \sqrt {{x_2} - 3} }}{{{x_1} - {x_2}}}\\
= \frac{{{x_1} - 3 - {x_2} + 3}}{{({x_1} - {x_2})(\sqrt {{x_1} - 3} + \sqrt {{x_2} - 3} )}}\\
= \frac{{{x_1} - {x_2}}}{{({x_1} - {x_2})(\sqrt {{x_1} - 3} + \sqrt {{x_2} - 3} )}}\\
= \frac{1}{{\sqrt {{x_1} - 3} + \sqrt {{x_2} - 3} }} > 0
\end{array}\]

Vậy hám số đông biến

\[\begin{array}{l}
b)\;y = \sqrt {2 - x} + 2\sqrt {1 - x} \\
dk:x \le 1\\
Lay:{\rm{ }}{x_1}\; \ne {\rm{ }}{x_2}\; \in {\rm{ }}R,{\rm{ }}ta{\rm{ }}co:\\
\frac{{y({x_1}) - y({x_2})}}{{{x_1} - {x_2}}} = \frac{{\sqrt {2 - {x_1}} + 2\sqrt {1 - {x_1}} - (\sqrt {2 - {x_2}} + 2\sqrt {1 - {x_2}} )}}{{{x_1} - {x_2}}}\\
= \frac{{\sqrt {2 - {x_1}} - \sqrt {2 - {x_2}} + 2(\sqrt {1 - {x_1}} - \sqrt {1 - {x_2}} )}}{{{x_1} - {x_2}}}\\
= \frac{{2 - {x_1} - 2 + {x_2}}}{{({x_1} - {x_2})(\sqrt {2 - {x_1}} + \sqrt {2 - {x_2}} )}} + 2.\frac{{1 - {x_1} - 1 + {x_2}}}{{({x_1} - {x_2})(\sqrt {1 - {x_1}} + \sqrt {1 - {x_2}} )}}\\
= \frac{{{x_2} - {x_1}}}{{({x_1} - {x_2})(\sqrt {2 - {x_1}} + \sqrt {2 - {x_2}} )}} + 2.\frac{{{x_2} - {x_1}}}{{({x_1} - {x_2})(\sqrt {1 - {x_1}} + \sqrt {1 - {x_2}} )}}\\
= \frac{{ - 1}}{{\sqrt {2 - {x_1}} + \sqrt {2 - {x_2}} }} - \frac{2}{{\sqrt {1 - {x_1}} + \sqrt {1 - {x_2}} }} < 0
\end{array}\]

vậy hàm số nghịch biến

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Answer 2

\[\begin{array}{l}
a)y = {\rm{ }}f\left( x \right) = \sqrt {x - 3} \\
dkxd:x \ge 3\\
Lay:{\rm{ }}{x_1}\; \ne {\rm{ }}{x_2}\; \in {\rm{ }}R,{\rm{ }}ta{\rm{ }}co:\\
\frac{{f({x_1}) - f({x_2})}}{{{x_1} - {x_2}}} = \frac{{\sqrt {{x_1} - 3} - \sqrt {{x_2} - 3} }}{{{x_1} - {x_2}}}\\
= \frac{{{x_1} - 3 - {x_2} + 3}}{{({x_1} - {x_2})(\sqrt {{x_1} - 3} + \sqrt {{x_2} - 3} )}}\\
= \frac{{{x_1} - {x_2}}}{{({x_1} - {x_2})(\sqrt {{x_1} - 3} + \sqrt {{x_2} - 3} )}}\\
= \frac{1}{{\sqrt {{x_1} - 3} + \sqrt {{x_2} - 3} }} > 0
\end{array}\]

Vậy hám số đông biến

\[\begin{array}{l}
b)\;y = \sqrt {2 - x} + 2\sqrt {1 - x} \\
dk:x \le 1\\
Lay:{\rm{ }}{x_1}\; \ne {\rm{ }}{x_2}\; \in {\rm{ }}R,{\rm{ }}ta{\rm{ }}co:\\
\frac{{y({x_1}) - y({x_2})}}{{{x_1} - {x_2}}} = \frac{{\sqrt {2 - {x_1}} + 2\sqrt {1 - {x_1}} - (\sqrt {2 - {x_2}} + 2\sqrt {1 - {x_2}} )}}{{{x_1} - {x_2}}}\\
= \frac{{\sqrt {2 - {x_1}} - \sqrt {2 - {x_2}} + 2(\sqrt {1 - {x_1}} - \sqrt {1 - {x_2}} )}}{{{x_1} - {x_2}}}\\
= \frac{{2 - {x_1} - 2 + {x_2}}}{{({x_1} - {x_2})(\sqrt {2 - {x_1}} + \sqrt {2 - {x_2}} )}} + 2.\frac{{1 - {x_1} - 1 + {x_2}}}{{({x_1} - {x_2})(\sqrt {1 - {x_1}} + \sqrt {1 - {x_2}} )}}\\
= \frac{{{x_2} - {x_1}}}{{({x_1} - {x_2})(\sqrt {2 - {x_1}} + \sqrt {2 - {x_2}} )}} + 2.\frac{{{x_2} - {x_1}}}{{({x_1} - {x_2})(\sqrt {1 - {x_1}} + \sqrt {1 - {x_2}} )}}\\
= \frac{{ - 1}}{{\sqrt {2 - {x_1}} + \sqrt {2 - {x_2}} }} - \frac{2}{{\sqrt {1 - {x_1}} + \sqrt {1 - {x_2}} }} < 0
\end{array}\]

vậy hàm số nghịch biến

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