\(A=\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}\)

=\(\left(\sqrt{2}\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}\)

=\(\sqrt{2}\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}\)

=\(\left(\sqrt{3}+1\right)\sqrt{2}\sqrt{\sqrt{3}+2}\left(\sqrt{3}-2\right)\)

=\(\left(\sqrt{3}+1\right)\sqrt{2\left(\sqrt{3}+2\right)}\left(\sqrt{3}-2\right)\)

=\(\left(\sqrt{3}+1\right)\sqrt{2\sqrt{3}+4}\left(\sqrt{3}-2\right)\)

=\(\left(\sqrt{3}+1\right)\sqrt{3+2\sqrt{3}+1}\left(\sqrt{3}-2\right)\)

=\(\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\left(\sqrt{3}-2\right)\)

=\(\left(\sqrt{3}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\)

=\(\left(\sqrt{3}+1\right)^2\left(\sqrt{3}-2\right)\)

=\(\left(4+2\sqrt{3}\right)\left(\sqrt{3}-2\right)\)

=\(4\sqrt{3}-8+2\left(\sqrt{3}\right)^2-4\sqrt{3}\)

=\(-8+2.3\)

=\(-8+6=-2\)

19 100%

\(\frac{40}{2}:2=10\)

\(\frac{60}{3}:3=10\)

\(\frac{80}{4}:2=10\)

.........................

\(P=\frac{\sqrt{x}+\sqrt{y}}{xy}=\frac{\sqrt{x}+\sqrt{y}}{16}\ge\frac{2\sqrt{\sqrt{x}.\sqrt{y}}}{16}=\frac{\sqrt[4]{xy}}{8}=\frac{\sqrt[4]{16}}{8}=\frac{2}{8}=\frac{1}{4}\)

Dấu "=" xảy ra khi và chỉ khi \(\sqrt{x}=\sqrt{y}\text{ và }x.y=16\text{ }\Leftrightarrow x=y=4\)

Vậy GTNN của P là 1/4 khi x = y = 4

ĐỀ bài bài 2 là: Đưa thừa số vào dấu căn bậc hai

a/ đề \(=\frac{\sqrt{27}}{3}.\frac{\sqrt{81}}{\sqrt{3}}=\sqrt{3}.3\sqrt{3}=9\)

b/ \(=-\sqrt{10}+\sqrt{7}\)

c/ \(=\sqrt{5}-\sqrt{2}\)