Đáp án:
2b) x=4
Giải thích các bước giải:
\(\begin{array}{l}
C1:\\
1)A = \left| {\sqrt 5 - 1} \right| + 1 = \sqrt 5 - 1 + 1 = \sqrt 5 \\
B = \left( {2\sqrt 3 + 4\sqrt 5 } \right).\sqrt 3 - 2\sqrt {15} \\
= 6 + 4\sqrt {15} - 2\sqrt {15} \\
= 6 + 2\sqrt {15} \\
2)a)P = \dfrac{{2\sqrt x + \sqrt x + 1 + \sqrt x \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{3\sqrt x + 1 + x - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + 2\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
b)P = 3\\
\to \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}} = 3\\
\to \sqrt x + 1 = 3\sqrt x - 3\\
\to 2\sqrt x = 4\\
\to \sqrt x = 2\\
\to x = 4
\end{array}\)
