\(\sqrt{27+10\sqrt{2}}-3\sqrt{2}\)

Rút gọn

">
K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

=5+căn 2-3căn 2

=5-2căn 2

26 tháng 6 2023

Latex đi anh.

21 tháng 9 2017

\(\left(\sqrt{12}+2\sqrt{27}-\sqrt{3}\right):\sqrt{3}\)

\(=\sqrt{12}:\sqrt{3}+2\sqrt{27}:\sqrt{3}-\sqrt{3}:\sqrt{3}\)

\(=\sqrt{4}+2\sqrt{9}-1\)

\(=2+6-1\)

\(=7\)

21 tháng 9 2017

2) \(\left(4\sqrt{2}-\sqrt{8}+2\right).\sqrt{2-\sqrt{8}}\)

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

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

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

\(=12.\left(2-2\sqrt{2}\right)\)

\(=24-24\sqrt{2}\)

\(=24\left(1-\sqrt{2}\right)\)

3) \(\sqrt{3}\left(2\sqrt{27}-\sqrt{75}+\frac{3}{2}\sqrt{12}\right)\)

\(=\sqrt{3}\left(2\sqrt{3^2.3}-\sqrt{5^2.3}+\frac{3}{2}\sqrt{2^2.3}\right)\)

\(=\sqrt{3}\left(6\sqrt{3}-5\sqrt{3}+3\sqrt{3}\right)\)

\(=\sqrt{3}.4\sqrt{3}\)

\(=12\)

20 tháng 10 2018

 a) \(\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{7\left(\sqrt{3}+\sqrt{5}\right)}}=\) \(\frac{\sqrt{2}}{\sqrt{7}}\)

 b ) \(\frac{15\sqrt{2}+9\sqrt{3}}{3\sqrt{3}+3\sqrt{5}}=\frac{3\left(5\sqrt{2}+3\sqrt{3}\right)}{3\left(\sqrt{3}+\sqrt{5}\right)}\)\(=\frac{5\sqrt{2}+3\sqrt{3}}{\sqrt{3}+\sqrt{5}}\)

c)\(\frac{\sqrt{2}-\sqrt{6}+\sqrt{3}-\sqrt{9}+\sqrt{4}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\) =  \(\frac{\sqrt{2}\left(1-\sqrt{3}\right)+\sqrt{3}\left(1-\sqrt{3}\right)+\sqrt{4}\left(1-\sqrt{3}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)\(=\frac{\left(1-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1-\sqrt{3}\)

 d) \(\frac{\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{5}-1}=\frac{\sqrt{5}-1}{\sqrt{5}-1}=1\)

\(\sqrt{300}-\sqrt{27}+4\sqrt{3}\)

=\(10\sqrt{3}-3\sqrt{3}+4\sqrt{3}\)

=\(11\sqrt{3}\)

25 tháng 7 2019

\(\sqrt{300}-\sqrt{27}+4\sqrt{3}\)

\(=\sqrt{10^2.3}-\sqrt{3^2.3}+4\sqrt{3}\)

\(=10\sqrt{3}-3\sqrt{3}+4\sqrt{3}\)

\(=11\sqrt{3}\)

\(\frac{2+2\sqrt{5}}{3-\sqrt{5}}=\frac{\left(2+2\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}=\frac{6+2\sqrt{5}+6\sqrt{5}+10}{3^2-\sqrt{5}^2}=\frac{16+8\sqrt{5}}{4}=\frac{4\left(4+2\sqrt{5}\right)}{4}=4+2\sqrt{5}\)

13 tháng 8 2018

\(B=\frac{9\sqrt{5}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}=\frac{9\sqrt{5}+9\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{9\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}=9\)

\(C=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{4}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

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

mik chỉnh lại đề

\(D=\frac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}=\frac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}\)

\(=\frac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}{3\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}=\frac{2}{3}\)

11 tháng 5 2024

$\dfrac{\sqrt{3}}{8}a^3$.

18 tháng 9 2019

d/ \(x=\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)

\(\Leftrightarrow x^3=3+\sqrt{9+\frac{125}{27}}+3-\sqrt{9+\frac{125}{27}}-3\left(\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\right)\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}.\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)

\(\Leftrightarrow x^3=6-3x\sqrt[3]{9-9-\frac{125}{27}}\)

\(\Leftrightarrow x^3=6-5x\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+6\right)=0\)

\(\Leftrightarrow x=1\)

19 tháng 9 2019

c/

\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)

\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)

\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{12}+4}}}\)

\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)

\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)

\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)

\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)

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

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

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

\(=3-1=2\)

3 tháng 6 2016

4.828587429