Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) = \(5\sqrt{2}-3\sqrt{6}+3\sqrt{2}+5\sqrt{6}\)
= \(8\sqrt{2}+2\sqrt{6}\)
b) = \(2\sqrt{3}-4\sqrt{2}-5\sqrt{3}-\sqrt{2}\)
= \(-3\sqrt{3}-5\sqrt{2}\)
c) = \(\frac{\left(\sqrt{2}-1\right)\left(2+\sqrt{2}\right)}{\left(2-\sqrt{2}\right)\left(2+\sqrt{2}\right)}\)
=\(\frac{2\sqrt{2}+2-2-\sqrt{2}}{2^2-\sqrt{2^2}}\)
=\(\frac{\sqrt{2}}{4-2}\) = \(\frac{\sqrt{2}}{2}\)
d) = \(2\sqrt{6}-5\sqrt{6}+2\sqrt{2}\)
=\(-3\sqrt{6}+2\sqrt{2}\)
e) = \(8\sqrt{6}+3\sqrt{6}-6\sqrt{6}=5\sqrt{6}\)
f) = \(4\sqrt{3}+9\sqrt{3}-4\sqrt{3}=9\sqrt{3}\)
g) = \(10+5\sqrt{10}-5\sqrt{10}=10\)
h) = \(\frac{\left(3+\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}+\frac{\left(3-\sqrt{3}\right)\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
= \(\frac{9+3\sqrt{3}+3\sqrt{3}+3}{3^2-\sqrt{3^2}}+\frac{9-3\sqrt{3}-3\sqrt{3}+3}{3^2-\sqrt{3^2}}\)
= \(\frac{12+6\sqrt{3}}{9-3}+\frac{12-6\sqrt{3}}{9-3}\)
= \(\frac{12+6\sqrt{3}+12-6\sqrt{3}}{6}\)
= \(\frac{24}{6}=4\)
k) = \(\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right).\sqrt{7}+2\sqrt{21}\)
= \(\left(3\sqrt{7}-2\sqrt{3}\right).\sqrt{7}+2\sqrt{21}\)
= \(21-2\sqrt{21}+2\sqrt{21}=21\)
l) = \(\frac{\left(2\sqrt{3}-\sqrt{6}\right)\left(\sqrt{8}+2\right)}{\left(\sqrt{8}-2\right)\left(\sqrt{8}+2\right)}\)
= \(\frac{4\sqrt{6}+4\sqrt{3}-4\sqrt{3}-2\sqrt{6}}{\sqrt{8^2}-2^2}\)
= \(\frac{2\sqrt{6}}{8-4}=\frac{2\sqrt{6}}{4}=\frac{\sqrt{6}}{2}\)
a) Ta có: \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=\left(-\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=-2+2\sqrt{5}-\sqrt{5}\)
\(=-2+\sqrt{5}\)
b) \(\left(\frac{1}{2}\sqrt{\frac{1}{2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\sqrt{200}\right)\div\frac{1}{8}\)
\(=\left(\frac{\sqrt{2}}{4}-\frac{3\sqrt{2}}{2}+8\sqrt{2}\right)\cdot8\)
\(=\frac{27\sqrt{2}}{4}\cdot8\)
\(=54\sqrt{2}\)
a, \(\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)^2}\) +\(\frac{1}{\left(\sqrt{3}-\sqrt{2}\right)^2}\) =\(\frac{\left(\sqrt{3}+\sqrt{2}\right)^2+\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\frac{10}{1}=10\)
mấy câu còn lại bạn tự làm nốt nhé mk ban rồi
a)
\(4\sqrt{7}=\sqrt{4^2.7}=\sqrt{112}\)
\(3\sqrt{13}=\sqrt{3^2.13}=\sqrt{117}\)
\(\sqrt{112}< \sqrt{117}\Rightarrow 4\sqrt{7}< 3\sqrt{13}\)
b) \(3\sqrt{12}=\sqrt{3^2.12}=\sqrt{9.2^2.3}=2\sqrt{27}>2\sqrt{16}\)
c)
\(\frac{1}{4}\sqrt{82}=\sqrt{\frac{82}{16}}=\sqrt{\frac{41}{8}}=\sqrt{5+\frac{1}{8}}\)
\(6\sqrt{\frac{1}{7}}=\sqrt{\frac{36}{7}}=\sqrt{5+\frac{1}{7}}\)
\(\sqrt{5+\frac{1}{8}}< \sqrt{5+\frac{1}{7}}\Rightarrow \frac{1}{4}\sqrt{82}< 6\sqrt{\frac{1}{7}}\)
d)
\(\frac{1}{2}\sqrt{\frac{17}{2}}=\sqrt{\frac{17}{8}}=\sqrt{2+\frac{1}{8}}\)
\(\frac{1}{3}\sqrt{19}=\sqrt{\frac{19}{9}}=\sqrt{2+\frac{1}{9}}\)
\(\sqrt{2+\frac{1}{8}}>\sqrt{2+\frac{1}{9}}\Rightarrow \frac{1}{2}\sqrt{\frac{17}{2}}> \frac{1}{3}\sqrt{19}\)
e)
\(3\sqrt{3}-2\sqrt{2}=\sqrt{27}-\sqrt{8}\)
Mà \(\sqrt{27}>\sqrt{25}; \sqrt{8}< \sqrt{9}\Rightarrow \sqrt{27}-\sqrt{8}> \sqrt{25}-\sqrt{9}=5-3=2\)
Vậy \(3\sqrt{3}-2\sqrt{2}>2\)
f)
\(\sqrt{7}+\sqrt{5}< \sqrt{9}+\sqrt{9}=6\)
\(\sqrt{49}=7\)
\(\Rightarrow \sqrt{7}+\sqrt{5}< 6< 7=\sqrt{49}\)
g)
\(\sqrt{2}< \sqrt{3}; \sqrt{11}< \sqrt{25}=5\)
\(\Rightarrow \sqrt{2}+\sqrt{11}< \sqrt{3}+5\)
h) Lặp lại câu d
i)
\(\sqrt{21}>\sqrt{20}\); \(\sqrt{5}< \sqrt{6}\)
\(\Rightarrow \sqrt{21}-\sqrt{5}> \sqrt{20}-\sqrt{6}\)
\(\sqrt{50}-\sqrt{54}+\frac{1}{2}\times\sqrt{72}+\frac{5}{6}\times\sqrt{216}\)
\(=\sqrt{25.5}-\sqrt{9.6}+\frac{1}{2}\times\sqrt{36.2}+\frac{5}{6}\times\sqrt{36.6}\)
\(=5\sqrt{5}-3\sqrt{6}+\frac{1}{2}\times6\sqrt{2}+\frac{5}{6}\times6\sqrt{6}\)
\(=5\sqrt{5}-3\sqrt{6}+3\sqrt{2}+5\sqrt{6}\)
\(=5\sqrt{5}+2\sqrt{6}+3\sqrt{2}\)
k mk nha
thank you