Đáp án:
\(\begin{array}{l}
C1:\\
B = 2\\
A = \sqrt 5 + 1\\
C = 3\sqrt 5 \\
D = 6 \\
P = - 1\\
C2:\\
B = 5\sqrt 6 \\
A = \sqrt 5 \\
C3:\\
A = \dfrac{1}{{x - 1}}\\
C4:\\
1)\dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
2)x = \dfrac{1}{4}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
C1:\\
B = \left| {2 - \sqrt 3 } \right| + \sqrt 3 \\
= 2 - \sqrt 3 + \sqrt 3 = 2\\
A = \sqrt {5 + 2\sqrt 5 .1 + 1} = \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} \\
= \left| {\sqrt 5 + 1} \right| = \sqrt 5 + 1\\
C = \sqrt {5 - 2.\sqrt 5 .\sqrt 2 + 2} + 2\sqrt 5 + \dfrac{1}{2}.2\sqrt 2 \\
= \sqrt {{{\left( {\sqrt 5 - \sqrt 2 } \right)}^2}} + 2\sqrt 5 + \sqrt 2 \\
= \sqrt 5 - \sqrt 2 + 2\sqrt 5 + \sqrt 2 \\
= 3\sqrt 5 \\
D = \sqrt 2 \left( {3 + \sqrt 3 } \right).\sqrt {6 - 3\sqrt 3 } \\
= \left( {3 + \sqrt 3 } \right).\sqrt {12 - 6\sqrt 3 } \\
= \left( {3 + \sqrt 3 } \right).\sqrt {9 - 2.3.\sqrt 3 + 3} \\
= \left( {3 + \sqrt 3 } \right).\sqrt {{{\left( {3 - \sqrt 3 } \right)}^2}} \\
= \left( {3 + \sqrt 3 } \right).\left( {3 - \sqrt 3 } \right)\\
= 9 - 3=6 \\
P = \dfrac{{\sqrt {3 - 2\sqrt 3 .1 + 1} }}{{1 - \sqrt 3 }}\\
= \dfrac{{\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} }}{{1 - \sqrt 3 }}\\
= \dfrac{{\sqrt 3 - 1}}{{ - \left( {\sqrt 3 - 1} \right)}} = - 1\\
C2:\\
B = \dfrac{{2\left( {\sqrt 7 + \sqrt 6 } \right)}}{{7 - 6}} - 2\sqrt 7 + 3\sqrt 6 \\
= 2\sqrt 7 + 2\sqrt 6 - 2\sqrt 7 + 3\sqrt 6 \\
= 5\sqrt 6 \\
A = \dfrac{{\left( {5 + \sqrt 5 } \right)\left( {\sqrt 5 - 2} \right)}}{{5 - 4}} + \dfrac{{\sqrt 5 \left( {\sqrt 5 + 1} \right)}}{{5 - 1}} - \dfrac{{3\sqrt 5 \left( {3 - \sqrt 5 } \right)}}{{9 - 5}}\\
= \left( {5 + \sqrt 5 } \right)\left( {\sqrt 5 - 2} \right) + \dfrac{{\sqrt 5 \left( {\sqrt 5 + 1} \right)}}{4} - \dfrac{{3\sqrt 5 \left( {3 - \sqrt 5 } \right)}}{4}\\
= \dfrac{{4\left( {5\sqrt 5 - 10 + 5 - 2\sqrt 5 } \right) + 5 + \sqrt 5 - 9\sqrt 5 + 15}}{4}\\
= \dfrac{{12\sqrt 5 - 20 + 5 + \sqrt 5 - 9\sqrt 5 + 15}}{4}\\
= \dfrac{{4\sqrt 5 }}{4} = \sqrt 5 \\
C3:\\
A = \dfrac{{2\sqrt x + x - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}:\dfrac{{x + \sqrt x + 1 - \sqrt x - 2}}{{x + \sqrt x + 1}}\\
= \dfrac{{\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{x + \sqrt x + 1}}{{x - 1}}\\
= \dfrac{1}{{x - 1}}\\
C4:\\
1)Q = \dfrac{{\sqrt x + 1 - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\left[ {\dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}} - \dfrac{{1 - \sqrt x }}{{\sqrt x \left( {1 - \sqrt x } \right)}}} \right]\\
= \dfrac{{\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\left( {\sqrt x - \dfrac{1}{{\sqrt x }}} \right)\\
= \dfrac{1}{{\sqrt x + 1}}.\dfrac{{x - 1}}{{\sqrt x }}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
2)Q = - 1\\
\to \dfrac{{\sqrt x - 1}}{{\sqrt x }} = - 1\\
\to \sqrt x - 1 = - \sqrt x \\
\to 2\sqrt x = 1\\
\to \sqrt x = \dfrac{1}{2}\\
\to x = \dfrac{1}{4}
\end{array}\)
