Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
câu E dễ nhất nên mình làm trước , các câu còn lại làm tương tự ( biến đổi thành hằng đẳng thức rồi rút gọn ) :
\(E=\sqrt{9-2.3.\sqrt{6}+6}+\sqrt{24-2.2\sqrt{6}.3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3\) ( vì \(3-\sqrt{6}>0;2\sqrt{6}-3>0\) )
\(=\sqrt{6}\)
a) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
= \(2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(2-\sqrt{3}+\sqrt{3}-1\) = \(1\)
b) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
= \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3-2\sqrt{6}\right)^2}\)
= \(3-\sqrt{6}+2\sqrt{6}-3\) = \(\sqrt{6}\)
c) \(\left(15\sqrt{200}-3\sqrt{450}+2\sqrt{50}\right):\sqrt{10}\)
= \(\dfrac{15\sqrt{200}}{\sqrt{10}}-\dfrac{3\sqrt{450}}{\sqrt{10}}+\dfrac{2\sqrt{50}}{\sqrt{10}}\)
= \(15\sqrt{20}-3\sqrt{45}+2\sqrt{5}\)
= \(30\sqrt{5}-9\sqrt{5}+2\sqrt{5}\) = \(23\sqrt{5}\)
\(\sqrt{10-4\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{2^2-2.2.\sqrt{6}+\left(\sqrt{6}\right)^2}+\sqrt{3^2-2.3.2\sqrt{6}+\left(2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{6}\right)^2}+\sqrt{\left(3-2\sqrt{6}\right)^2}\)
\(=-\left(2-\sqrt{6}\right)-\left(3-2\sqrt{6}\right)\)
\(=-2+\sqrt{6}-3+2\sqrt{6}\)
\(=-5+3\sqrt{6}\)
\(\sqrt{16-6\sqrt{7}}+\sqrt{32-8\sqrt{7}}\)
\(=\sqrt{3^2-2.3.\sqrt{7}+\left(\sqrt{7}\right)^2}+\sqrt{2^2-2.2.2\sqrt{7}+\left(2\sqrt{7}\right)^2}\)
\(=\sqrt{\left(3-\sqrt{7}\right)^2}+\sqrt{\left(2-2\sqrt{7}\right)^2}\)
\(=3-\sqrt{7}-\left(2-2\sqrt{7}\right)\)
\(=3-\sqrt{7}-2+2\sqrt{7}\)
\(=1+\sqrt{7}\)
B=\(\frac{6-6\sqrt{3}}{1-\sqrt{3}}+\frac{3\sqrt{3}+3}{\sqrt{3}+1}=\frac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\frac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
C=\(\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}=\frac{3\left(1+\sqrt{3}\right)}{\sqrt{3}}+\frac{\sqrt{3}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}=\sqrt{3}+1-\sqrt{3}=1\)
D=\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
E=\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\sqrt{3}+\frac{1}{2-\sqrt{3}}=\frac{2\sqrt{3}-1}{2-\sqrt{3}}\)
mình làm mẫu 2 bài nhé 2 bài kia bạn làm tương tự
1)a)\(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{3}=\sqrt{3}+1-\sqrt{3}=1\)
\(\sqrt{10-2\sqrt{21}}+\sqrt{7}=\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}+\sqrt{7}=\sqrt{7}+\sqrt{3}+\sqrt{7}=2\sqrt{7}+\sqrt{3}\)
2)a) \(\sqrt{12-6\sqrt{3}}-\sqrt{3}=\sqrt{\left(3-\sqrt{3}\right)^2}-\sqrt{3}=3-\sqrt{3}-\sqrt{3}=3-2\sqrt{3}\)
b) \(\sqrt{7+2\sqrt{6}}-\sqrt{3}=\sqrt{\left(1+\sqrt{6}\right)^2}-\sqrt{3}=1+\sqrt{6}-\sqrt{3}\)
\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}=2-\sqrt{3}+\sqrt{3-2\sqrt{3}+1}\left(\text{vì }2>\sqrt{3}\right)\)
\(=2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}=2-\sqrt{3}+\sqrt{3}-1\left(\text{vì }\sqrt{3}>1\right)\)
\(=1\)
\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{9-2.3\sqrt{6}+6}+\sqrt{\left(2\sqrt{6}\right)^2-2.2\sqrt{6}.3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}=3-\sqrt{6}+2\sqrt{6}-3\left(\text{vì }3>\sqrt{6};2\sqrt{6}>3\right)\)
\(=\sqrt{6}\)
a/
\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=2-\sqrt{3}+\sqrt{3}-1\)
\(=1\)
b/
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(\sqrt{24}-3\right)^2}\)
\(=3-\sqrt{6}+\sqrt{24}-3\)
\(=-\sqrt{6}+2\sqrt{6}=\sqrt{6}\)
tick cho mình nha
1. \(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\\ =\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}=2\sqrt{2}\)
2. \(\sqrt{12-6\sqrt{3}}+\sqrt{21-12\sqrt{3}}=\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{3}-3\right)^2}\\ =3-\sqrt{3}+2\sqrt{3}-3=\sqrt{3}\)
3. \(\sqrt{33-12\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(3+\sqrt{6}\right)^2}\\ =2\sqrt{6}-3+3+\sqrt{6}=3\sqrt{6}\)
1.\(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\)
\(=\sqrt{5}+\sqrt{2}-\left(\sqrt{5}-\sqrt{2}\right)=2\sqrt{2}\)
2. \(\sqrt{12-6\sqrt{3}+\sqrt{21-12\sqrt{3}}}=\sqrt{12-6\sqrt{3}+\sqrt{\left(3-2\sqrt{3}\right)^2}}\)
\(=\sqrt{12-6\sqrt{3}+2\sqrt{3}-3}=\sqrt{9-4\sqrt{3}}\)
3. \(\sqrt{33-12\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(\sqrt{6}-3\right)^2}\)
\(=2\sqrt{6}-3+3-\sqrt{6}=\sqrt{6}\)