Đáp án:
B3:
c. 2
Giải thích các bước giải:
\(\begin{array}{l}
B2:\\
d.\dfrac{{4\left( {3 + \sqrt 5 } \right)}}{{9 - 5}} - \dfrac{{3\left( {\sqrt 5 - \sqrt 2 } \right)}}{{5 - 2}} - \dfrac{{\sqrt 2 + 1}}{{2 - 1}}\\
= \dfrac{{12 + 4\sqrt 5 }}{4} - \dfrac{{3\sqrt 5 - 3\sqrt 2 }}{3} - \dfrac{{\sqrt 2 + 1}}{1}\\
= \dfrac{{36 + 12\sqrt 5 - 12\sqrt 5 + 12\sqrt 2 - 12\sqrt 2 - 12}}{{12}}\\
= \dfrac{{24}}{{12}} = 2\\
f.\dfrac{{3\left( {\sqrt 7 + 2} \right)}}{{7 - 4}} + \dfrac{{4\left( {\sqrt {11} - \sqrt 7 } \right)}}{{11 - 7}} - \dfrac{{2\left( {\sqrt {11} + 3} \right)}}{{11 - 9}}\\
= \dfrac{{3\sqrt 7 + 6}}{3} + \dfrac{{4\sqrt {11} - 4\sqrt 7 }}{4} - \dfrac{{2\sqrt {11} + 6}}{2}\\
= \dfrac{{12\sqrt 7 + 24 + 12\sqrt {11} - 12\sqrt 7 - 12\sqrt {11} - 36}}{{12}}\\
= \dfrac{{ - 12}}{{12}} = - 1\\
B3:\\
a.\sqrt {8 + 2\sqrt {15} } - \sqrt {8 - 2\sqrt {15} } \\
= \sqrt {5 + 2\sqrt 5 .\sqrt 3 + 3} - \sqrt {5 - 2\sqrt 5 .\sqrt 3 + 3} \\
= \sqrt {{{\left( {\sqrt 5 + \sqrt 3 } \right)}^2}} - \sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} \\
= \sqrt 5 + \sqrt 3 - \sqrt 5 + \sqrt 3 = 2\sqrt 3 \\
b.\sqrt {12 + 6\sqrt 3 } - \sqrt {4 - 2\sqrt 3 } \\
= \sqrt {9 + 2.3.\sqrt 3 + 3} - \sqrt {3 - 2\sqrt 3 .1 + 1} \\
= \sqrt {{{\left( {3 + \sqrt 3 } \right)}^2}} - \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} \\
= 3 + \sqrt 3 - \sqrt 3 + 1 = 4\\
c.\sqrt {x + 1 + 2\sqrt x } + \sqrt {x + 1 - 2\sqrt x } \\
= \sqrt {x + 2\sqrt x .1 + 1} + \sqrt {x - 2\sqrt x .1 + 1} \\
= \sqrt {{{\left( {\sqrt x + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt x - 1} \right)}^2}} \\
= \sqrt x + 1 + \left| {\sqrt x - 1} \right|\\
= \sqrt x + 1 - \left( {\sqrt x - 1} \right)\\
= 2\left( {do:0 \le x < 1} \right)
\end{array}\)
