Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\sqrt{47+\sqrt{5}}\cdot\sqrt{47-\sqrt{5}}\)
\(=\sqrt{2204}=2\sqrt{551}\)
\(B=5-2\sqrt{6}+10+\sqrt{6}=15-\sqrt{6}\)
\(A=\sqrt{\left(47+\sqrt{5}\right)\left(47-\sqrt{5}\right)}=2\sqrt{551}\)
\(B=5-2\sqrt{6}+10+\sqrt{6}=15-\sqrt{6}\)
B = \(\sqrt{\sqrt{75-2.2\sqrt{2}.5\sqrt{3}+8}+\sqrt{50-2.2\sqrt{3}.5\sqrt{2}+12}}.\sqrt{3\sqrt{3}-3\sqrt{2}}\)
= \(\sqrt{\sqrt{\left(5\sqrt{3}-2\sqrt{2}\right)^2}+\sqrt{\left(5\sqrt{2}+2\sqrt{3}\right)^2}}.\sqrt{3\sqrt{3}-3\sqrt{2}}\)
= \(\sqrt{5\sqrt{3}-2\sqrt{2}+5\sqrt{2}-2\sqrt{3}}.\sqrt{3\sqrt{3}-3\sqrt{2}}\)
= \(\sqrt{3\sqrt{3}+3\sqrt{2}}.\sqrt{3\sqrt{3}-3\sqrt{2}}=\sqrt{\left(3\sqrt{3}+3\sqrt{2}\right)\left(3\sqrt{3}-3\sqrt{2}\right)}\)
= \(\sqrt{27-18}=\sqrt{9}=3\)
Phân tích cái trong ngặc đầu thành: (5 căn 3 - 2 căn 2)^2
cái thứ 2 là ( 5 căn 2 - 2 căn 3)^2
sau đó phá đc 1 ngặc làm tiếp
a) \(\sqrt[4]{49+20\sqrt{6}}+\sqrt[4]{49-20\sqrt{6}}=\sqrt[4]{25+2\sqrt{600}+24}+\sqrt[4]{25-2\sqrt{600}+24}\\ =\sqrt[4]{\left(\sqrt{25}+\sqrt{24}\right)^2}+\sqrt[4]{\left(\sqrt{25}-\sqrt{24}\right)^2}=\sqrt{\sqrt{25}+\sqrt{24}}+\sqrt{\sqrt{25}-\sqrt{24}}\\ =\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{6}+2}+\sqrt{3-2\sqrt{6}+2}\\ =\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}\\ =2\sqrt{3}\)
\(A=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+....}}}}>0\)
\(\Rightarrow A^2=6+\sqrt{6+\sqrt{6+\sqrt{6+....}}}\)
\(\Rightarrow A^2=6+A\)\(\Rightarrow A^2-A-6=0\)
\(\Rightarrow\left(A-3\right)\left(A+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}A-3=0\\A+2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}A=3\\A=-3\end{cases}}\Rightarrow A=3>0\) (thỏa)
a. \(\sqrt{49-20\sqrt{6}}-\sqrt{106+20\sqrt{6}}=\sqrt{\left(5-2\sqrt{6}\right)^2}-\sqrt{\left(10+\sqrt{6}\right)^2}=5-2\sqrt{6}-10-\sqrt{6}=-5-3\sqrt{6}\)
b. \(\sqrt{83-20\sqrt{6}}+\sqrt{62-20\sqrt{6}}=\sqrt{\left(5\sqrt{3}-2\sqrt{2}\right)^2}+\sqrt{\left(5\sqrt{2}-2\sqrt{3}\right)^2}=5\sqrt{3}-2\sqrt{2}+5\sqrt{2}-2\sqrt{3}=3\sqrt{3}+3\sqrt{2}\)
c. \(\sqrt{302-20\sqrt{6}}+\sqrt{203-20\sqrt{6}}=\sqrt{\left(10\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(10\sqrt{2}-\sqrt{3}\right)^2}=10\sqrt{3}-\sqrt{2}+10\sqrt{2}-\sqrt{3}=9\sqrt{3}+9\sqrt{2}\)
d. \(\sqrt{601-20\sqrt{6}}-\sqrt{154-20\sqrt{6}}=\sqrt{\left(10\sqrt{6}-1\right)^2}-\sqrt{\left(5\sqrt{6}-2\right)^2}=10\sqrt{6}-1-5\sqrt{6}+2=1+5\sqrt{6}\)