Giải bài 1 trang 53 vở thực hành Toán 9

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

$=\sqrt{12}.\sqrt{12}+\sqrt{12}.\sqrt{3}=\sqrt{{12}^2}+\sqrt{36}=12+6=18;$

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

$=\sqrt{8}.\sqrt{50}-\sqrt{8}.\sqrt{2}=\sqrt{400}-\sqrt{16}=20-4=16;$

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

$={\left(\sqrt{3}\right)}^2+{\left(\sqrt{2}\right)}^2+2\sqrt{3}.\sqrt{2}-2\sqrt{6}=3+2=5$.