1. Calculate:

$$
(\sqrt{7+2 \sqrt{10}}+\sqrt{7-2 \sqrt{10}})^{2}
$$
First solution.

$(\sqrt{7+2 \sqrt{10}}+\sqrt{7-2 \sqrt{10}})^{2}=(\sqrt{7+2 \sqrt{10}}+\sqrt{7-2 \sqrt{10}}) \cdot(\sqrt{7+2 \sqrt{10}}+\sqrt{7-2 \sqrt{10}})$

$=(\sqrt{7+2 \sqrt{10}})^{2}+\sqrt{(7+2 \sqrt{10})(7-2 \sqrt{10})}+\sqrt{(7-2 \sqrt{10})(7+2 \sqrt{10})}+(\sqrt{7-2 \sqrt{10}})^{2}$

$=(\sqrt{7+2 \sqrt{10}})^{2}+2 \cdot \sqrt{(7+2 \sqrt{10})(7-2 \sqrt{10})}+(\sqrt{7-2 \sqrt{10}})^{2}$

2 POINTS

$=7+2 \sqrt{10}+2 \cdot \sqrt{49-14 \sqrt{10}+14 \sqrt{10}-40}+7-2 \sqrt{10}$

$=7+2 \sqrt{10}+2 \cdot \sqrt{49-40}+7-2 \sqrt{10}$

$=14+2 \cdot 3$

Note: The student can use expressions for the square of a binomial and the difference of squares when solving the problem.

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