Đáp án:
\(\begin{array}{l}
a,\\
{\left( {\sqrt 3 + \sqrt 2 } \right)^2}\\
b,\\
{\left( {\sqrt 3 + 1} \right)^2}\\
c,\\
{\left( {2 - \sqrt 3 } \right)^2}\\
d,\\
{\left( {2\sqrt 5 - 3} \right)^2}\\
e,\\
{\left( {\sqrt 5 - 1} \right)^2}\\
h,\\
{\left( {\sqrt x - \sqrt y } \right)^3}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
5 + 2\sqrt 6 = 3 + 2\sqrt 6 + 2 = {\sqrt 3 ^2} + 2.\sqrt 3 .\sqrt 2 + {\sqrt 2 ^2} = {\left( {\sqrt 3 + \sqrt 2 } \right)^2}\\
b,\\
4 + 2\sqrt 3 = 3 + 2\sqrt 3 + 1 = {\sqrt 3 ^2} + 2.\sqrt 3 .1 + {1^2} = {\left( {\sqrt 3 + 1} \right)^2}\\
c,\\
7 - 4\sqrt 3 = 4 - 4\sqrt 3 + 3 = {2^2} - 2.2.\sqrt 3 + {\sqrt 3 ^2} = {\left( {2 - \sqrt 3 } \right)^2}\\
d,\\
29 - 12\sqrt 5 = 20 - 12\sqrt 5 + 9 = {\left( {2\sqrt 5 } \right)^2} - 2.2\sqrt 5 .3 + {3^2} = {\left( {2\sqrt 5 - 3} \right)^2}\\
e,\\
6 - \sqrt {20} = 6 - \sqrt {{2^2}.5} = 6 - 2\sqrt 5 = 5 - 2\sqrt 5 + 1\\
= {\sqrt 5 ^2} - 2.\sqrt 5 .1 + {1^2} = {\left( {\sqrt 5 - 1} \right)^2}\\
h,\\
x\sqrt x - 3x\sqrt y + 3y\sqrt x - y\sqrt y \\
= {\sqrt x ^3} - 3.{\sqrt x ^2}.\sqrt y + 3.\sqrt x .{\sqrt y ^2} - {\sqrt y ^3}\\
= {\left( {\sqrt x - \sqrt y } \right)^3}
\end{array}\)
