Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\left(\frac{\sqrt{9}}{2}+\frac{\sqrt{1}}{2}-\sqrt{2}\right)\sqrt{2}\)
\(=\frac{3\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-2\)
\(=\frac{4\sqrt{2}}{2}-2=2\sqrt{2}-2\)
b) \(\left(\frac{\sqrt{8}}{3}-\sqrt{24}+\frac{\sqrt{50}}{3}\right)\sqrt{6}\)
\(=\frac{4\sqrt{3}}{3}-12+\frac{10\sqrt{3}}{3}\)
\(=\frac{14\sqrt{3}}{3}-12\)
c) \(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{1}\right)\) (đã sửa đề)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\sqrt{2}\)
\(=\left(3-1\right)\sqrt{2}\)
\(=2\sqrt{2}\)
d) \(\left(3\sqrt{2}+1\right)\left(\sqrt{3\sqrt{2}-1}\right)\)
\(=\sqrt{3\sqrt{2}+1}\cdot\left(\sqrt{3\sqrt{2}+1}\cdot\sqrt{3\sqrt{2}-1}\right)\)
\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{18-1}\)
\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{17}\)
...
\(\sqrt{3}< 2;\sqrt{3}>1\)
\(\left|\sqrt{3}-2\right|+\left|\sqrt{3}-1\right|\)
\(2-\sqrt{3}+\sqrt{3}-1\)
\(=1\)
\(b,\sqrt{5}< \sqrt{9}=3\)
\(\left|\sqrt{5}-3\right|-2\sqrt{5}+2\)
\(3-\sqrt{5}-2\sqrt{5}+2\)
\(5-3\sqrt{5}\)
2\(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
= \(14-\sqrt{84}+7-\sqrt{84}\)
= 21
= 3 - 2\(\sqrt{2}\)+ 3 + 2\(\sqrt{2}\)= 3 + 3 = 6
\(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}=\left|3-2\sqrt{2}\right|+\left|3+2\sqrt{2}\right|=3-2\sqrt{2}+3+2\sqrt{2}=6\)