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Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 1\\
P = \dfrac{1}{{\sqrt {x – 1} + \sqrt x }} + \dfrac{1}{{\sqrt {x – 1} – \sqrt x }} + \dfrac{{x\sqrt x + x}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt {x – 1} – \sqrt x }}{{x – 1 – x}} + \dfrac{{\sqrt {x – 1} + \sqrt x }}{{x – 1 – x}} + \dfrac{{x\left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt {x – 1} – \sqrt x }}{{ – 1}} + \dfrac{{\sqrt {x – 1} + \sqrt x }}{{ – 1}} + x\\
= – \sqrt {x – 1} + \sqrt x – \sqrt {x – 1} – \sqrt x + x\\
= x – 2\sqrt {x – 1} \\
b)P = x – 2\sqrt {x – 1} \\
= x – 1 + 1 – 2\sqrt {x – 1} \\
= {\left( {\sqrt {x – 1} } \right)^2} – 2\sqrt {x – 1} + 1\\
= {\left( {\sqrt {x – 1} – 1} \right)^2} \ge 0\forall x \ge 1\\
\Rightarrow P \ge 0\forall x \ge 1
\end{array}$
Vậy P luôn không âm với mọi x xác định