Đáp án:
B3:
điều phải chứng minh
\(\begin{array}{l}
B4:\\
a) - 5\\
b)\dfrac{{96}}{{17}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B3:\\
a)VT = \left( {1 + \sqrt 2 } \right).\sqrt {2 - 2\sqrt 2 .1 + 1} \\
= \left( {1 + \sqrt 2 } \right).\sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} \\
= \left( {1 + \sqrt 2 } \right).\left( {\sqrt 2 - 1} \right)\\
= 2 - 1 = 1 = VP\\
\to dpcm\\
b)VT = \left( {2 - \sqrt 3 } \right)\left( {\sqrt 3 + 1} \right)\sqrt {3 + 2\sqrt 3 .1 + 1} \\
= \left( {2 - \sqrt 3 } \right)\left( {\sqrt 3 + 1} \right)\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} \\
= \left( {2 - \sqrt 3 } \right){\left( {\sqrt 3 + 1} \right)^2}\\
= \left( {2 - \sqrt 3 } \right)\left( {4 + 2\sqrt 3 } \right)\\
= \left( {2 - \sqrt 3 } \right)2\left( {2 + \sqrt 3 } \right)\\
= 2\left( {4 - 3} \right) = 2 = VP\\
\to dpcm\\
c)VT = \left( {3 - \sqrt 5 } \right).\sqrt 2 \left( {\sqrt 5 + 1} \right)\sqrt {3 + \sqrt 5 } \\
= \left( {3 - \sqrt 5 } \right).\left( {\sqrt 5 + 1} \right)\sqrt {6 + 2\sqrt 5 } \\
= \left( {3 - \sqrt 5 } \right).\left( {\sqrt 5 + 1} \right)\sqrt {5 + 2\sqrt 5 .1 + 1} \\
= \left( {3 - \sqrt 5 } \right).\left( {\sqrt 5 + 1} \right)\sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} \\
= \left( {3 - \sqrt 5 } \right).{\left( {\sqrt 5 + 1} \right)^2}\\
= \left( {3 - \sqrt 5 } \right).\left( {6 + 2\sqrt 5 } \right)\\
= \left( {3 - \sqrt 5 } \right).2\left( {3 + \sqrt 5 } \right)\\
= 2\left( {9 - 5} \right) = 2.4 = 8 = VP\\
\to dpcm\\
B4:\\
a)\left( {3\sqrt 2 - 2\sqrt 3 + 2\sqrt 2 - 3\sqrt 3 } \right)\left( {3\sqrt 2 - 2\sqrt 3 - 2\sqrt 2 + 3\sqrt 3 } \right)\\
= \left( {5\sqrt 2 - 5\sqrt 3 } \right)\left( {\sqrt 2 + \sqrt 3 } \right)\\
= 5\left( {\sqrt 2 - \sqrt 3 } \right)\left( {\sqrt 2 + \sqrt 3 } \right)\\
= 5\left( {2 - 3} \right) = - 5\\
b)\left( {3\sqrt 8 - 6.\sqrt {\dfrac{1}{2}} - 2\sqrt {18} + 3\sqrt {50} } \right):\left( {\dfrac{1}{2}.\sqrt {24,5} - \sqrt {4,5} + \dfrac{3}{4}.\sqrt {12,5} } \right)\\
= \left( {6\sqrt 2 - 3\sqrt 2 - 6\sqrt 2 + 15\sqrt 2 } \right):\left( {\dfrac{1}{2}.\dfrac{{7\sqrt 2 }}{2} - \dfrac{{3\sqrt 2 }}{2} + \dfrac{3}{4}.\dfrac{{5\sqrt 2 }}{2}} \right)\\
= 12\sqrt 2 :\dfrac{{17\sqrt 2 }}{8} = \dfrac{{96}}{{17}}
\end{array}\)
