Giải bài tập 3.7 trang 51

Đề bài

Tính:

a) $\sqrt{12}.\left(\sqrt{12}+\sqrt{3}\right);$

b) $\sqrt{8}.\left(\sqrt{50}-\sqrt{2}\right);$

c) ${\left(\sqrt{3}+\sqrt{2}\right)}^2-2\sqrt{6}.$

Lời giải chi tiết

a) $\sqrt{12}.\left(\sqrt{12}+\sqrt{3}\right)$

$\begin{array}{c}=\sqrt{12}.\sqrt{12}+\sqrt{12}.\sqrt{3}\\ =\sqrt{{12}^2}+\sqrt{36}\\ =12+6\\ =18\end{array}$

b) $\sqrt{8}.\left(\sqrt{50}-\sqrt{2}\right)$

$\begin{array}{c}=\sqrt{8}.\sqrt{50}-\sqrt{8}.\sqrt{2}\\ =\sqrt{400}-\sqrt{16}\\ =20-4\\ =16\end{array}$

c) ${\left(\sqrt{3}+\sqrt{2}\right)}^2-2\sqrt{6}$

$\begin{array}{c}={\sqrt{3}}^2+2.\sqrt{3}.\sqrt{2}+{\sqrt{2}}^2-2\sqrt{6}\\ =3+2\sqrt{6}+2-2\sqrt{6}\\ =5\end{array}$