Đáp án:
\(\begin{array}{l}
a)2\sqrt x - 1\\
b)2\sqrt 2 + 1\\
c)x \in \emptyset \\
d)0 \le x < \dfrac{1}{4}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)P = \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}} + \dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x - 1}}\\
= \sqrt x - 1 + \sqrt x \\
= 2\sqrt x - 1\\
b)Thay:x = 3 + 2\sqrt 2 \\
= 2 + 2\sqrt 2 .1 + 1\\
= {\left( {\sqrt 2 + 1} \right)^2}\\
\to P = 2\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} - 1\\
= 2\left( {\sqrt 2 + 1} \right) - 1\\
= 2\sqrt 2 + 1\\
c)P = 1\\
\to 2\sqrt x - 1 = 1\\
\to 2\sqrt x = 2\\
\to \sqrt x = 1\\
\to x = 1\left( l \right)\\
\to x \in \emptyset \\
d)P < 0\\
\to 2\sqrt x - 1 < 0\\
\to \sqrt x < \dfrac{1}{2}\\
\to 0 \le x < \dfrac{1}{4}
\end{array}\)
