Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
b/ \(\frac{2\sqrt{2}-1}{\sqrt{2}-1}+\frac{3\sqrt{2}-2}{\sqrt{2}-3}=\frac{\left(2\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}{1}+\frac{\left(2-3\sqrt{2}\right)\left(3+\sqrt{2}\right)}{\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)}\)
\(=3+\sqrt{2}+\frac{-7\sqrt{2}}{7}=3\)
c/ \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)
\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}=\sqrt{43+30\sqrt{2}}=\sqrt{\left(5+3\sqrt{2}\right)^2}=5+3\sqrt{2}\)
Mình đưa ra đáp án thôi nhé :)
a/ \(\left(\sqrt{\frac{5}{3}-\sqrt{\frac{3}{5}}}\right).\sqrt{15}=\sqrt{25-3\sqrt{15}}\)
b/ \(\frac{2\sqrt{2}-1}{\sqrt{2}-1}+\frac{3\sqrt{2}-2}{\sqrt{2}-3}=3\)
c/ \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=5+3\sqrt{2}\)
b)\(\sqrt{m+2\sqrt{m-1}}+\sqrt{m-2\sqrt{m-1}}\)
\(\Leftrightarrow\sqrt{m-1+2\sqrt{m-1}+1}+\sqrt{m-1-2\sqrt{m-1}+1}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{m-1}+1\right)^2}+\sqrt{\left(\sqrt{m-1}-1\right)^2}\)
\(\Leftrightarrow\sqrt{m-1}+1+\sqrt{m-1}-1\Leftrightarrow2\sqrt{m-1}\)
Câu 1 phá từng lớp ra :VD\(9+4\sqrt{2}\) =\((\sqrt{2}+2)^2\)
Câu 2:m+2\(\sqrt{m-1}\) =m-1+1+2\(\sqrt{m-1}\) =\((\sqrt{m-1} -1)^2 \)
b: \(=\dfrac{\sqrt{5}+1}{\sqrt{5}-1}+\dfrac{\sqrt{5}-1}{\sqrt{5}+1}\)
\(=\dfrac{6+2\sqrt{5}+6-2\sqrt{5}}{4}=\dfrac{12}{4}=3\)
c: \(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\)
e: \(=\dfrac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{\sqrt{2}\cdot\sqrt{3+\sqrt{3}-1}}{\sqrt{3}-1}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}-1}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(=\dfrac{4-2\sqrt{3}}{2}=2-\sqrt{3}\)
1. \(=\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}+\sqrt{3+\sqrt{\left(2\sqrt{3}+1\right)^2}}=\sqrt{5-2\sqrt{3}-1}+\sqrt{3+2\sqrt{3}+1}=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)
1/ \(\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\)
\(=\sqrt{5-\left(1+\sqrt{12}\right)^2}+\sqrt{3+\left(1+\sqrt{12}\right)^2}\)
\(=\sqrt{5-\left|1+\sqrt{12}\right|}+\sqrt{3+\left|1+\sqrt{12}\right|}\)
\(=\sqrt{5-1-\sqrt{12}}+\sqrt{3+1+\sqrt{12}}\)
\(=\sqrt{4-\sqrt{12}}+\sqrt{4+\sqrt{12}}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left|\sqrt{3}-1\right|+\left|\sqrt{3}+1\right|\)
\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)
a/ \(\sqrt{2}+\sqrt{6}\)
b/ Sửa đề:
\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}=1\)
c/ \(1+\sqrt{2}+\sqrt{5}\)
\(\text{a) }\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\\ =\sqrt{13+30\sqrt{2+\sqrt{8+1+4\sqrt{2}}}}\\ =\sqrt{13+30\sqrt{2+\sqrt{\left(\sqrt{8}+1\right)^2}}}\\ =\sqrt{13+30\sqrt{2+\sqrt{8}+1}}\\ =\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\\ =\sqrt{13+30\sqrt{2}+30}\\ =\sqrt{43+30\sqrt{2}}\\ =\sqrt{25+18+30\sqrt{2}}\\ =\sqrt{\left(5+\sqrt{18}\right)^2}\\ =5+3\sqrt{2}\)
\(\text{b) }\sqrt{m+2\sqrt{m-1}}+\sqrt{m-2\sqrt{m-1}}\\ =\sqrt{m-1+2\sqrt{m-1}+1}+\sqrt{m-1-2\sqrt{m-1}+1}\\ =\sqrt{\left(\sqrt{m-1}+1\right)^2}+\sqrt{\left(\sqrt{m-1}-1\right)^2}\\ =\sqrt{m-1}+1+\sqrt{m-1}-1\\ =2\sqrt{m-1}\)