Đáp án:
\(\begin{array}{l}
B1.2)\\
a)A = 1\\
B = 2\sqrt x - 1\\
b)x = 1\\
B1.3)\\
a)A = 1 - \sqrt x \\
B = 2\\
b)x = 9
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1.2)\\
a)A = \dfrac{{\sqrt 5 \left( {\sqrt 5 - 2} \right)}}{{\sqrt 5 }} - 5\sqrt 5 + 3 + 4\sqrt 5 \\
= \sqrt 5 - 2 - \sqrt 5 + 3 = 1\\
B = \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x }} + \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\sqrt x + 2}}\\
= \sqrt x + 1 + \sqrt x - 2 = 2\sqrt x - 1\\
b)A = B\\
\to 2\sqrt x - 1 = 1\\
\to 2\sqrt x = 2\\
\to \sqrt x = 1\\
\to x = 1\\
B1.3)\\
a)A = \dfrac{{1 + 2\sqrt x + x - 4\sqrt x }}{{1 - \sqrt x }}\\
= \dfrac{{1 - 2\sqrt x + x}}{{1 - \sqrt x }}\\
= \dfrac{{{{\left( {1 - \sqrt x } \right)}^2}}}{{1 - \sqrt x }} = 1 - \sqrt x \\
B = \left[ {\dfrac{{\sqrt 3 \left( {\sqrt 3 + 2} \right)}}{{\sqrt 3 }} + \dfrac{{\sqrt 2 \left( {\sqrt 2 + 1} \right)}}{{\sqrt 2 + 1}}} \right] - \dfrac{1}{{\sqrt 3 - \sqrt 2 }}\\
= \left( {\sqrt 3 + 2 + \sqrt 2 } \right) - \dfrac{{\sqrt 3 + \sqrt 2 }}{{3 - 2}}\\
= \sqrt 3 + 2 + \sqrt 2 - \sqrt 3 - \sqrt 2 = 2\\
b)A = B - 4\\
\to 1 - \sqrt x = 2 - 4\\
\to 1 - \sqrt x = - 2\\
\to \sqrt x = 3\\
\to x = 9
\end{array}\)
