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\(A=\sqrt{8}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =2\sqrt{2}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =4\sqrt{2}+4\sqrt{7}\)
\(B=\left(3+2\sqrt{6}+2\right)\left(25-20\sqrt{6}+24\right)\sqrt{3-2\sqrt{6}+2}\\ =\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}-\sqrt{2}\right)\\ =\left(\sqrt{3}+\sqrt{2}\right)\left(3-2\sqrt{6}+2\right)^2\\ =\left(\sqrt{3}-\sqrt{2}\right)^3\\ =9\sqrt{3}-11\sqrt{2}\)
a: \(A=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(\sqrt{2}\cdot B=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)
\(\Leftrightarrow B\sqrt{2}=3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}\)
\(\Leftrightarrow B\sqrt{2}=4\sqrt{5}\)
hay \(B=2\sqrt{10}\)
d: \(D\sqrt{2}=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\cdot\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5}-2\sqrt{5}+2=2\)
hay \(D=\sqrt{2}\)
\(\left(3\sqrt{2}+\sqrt{6}\right)\left(6-3\sqrt{3}\right)\)
\(=\sqrt{6}\left(\sqrt{3}+1\right)\times3\left(2-\sqrt{3}\right)\)
\(=\dfrac{3\sqrt{6}}{2}\left(\sqrt{3}+1\right)\left(4-2\sqrt{3}\right)\)
\(=\dfrac{3\sqrt{6}}{2}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)^2\)
\(=\dfrac{3\sqrt{6}}{2}\left(3-1\right)\left(\sqrt{3}-1\right)\)
\(=3\sqrt{6}\left(\sqrt{3}-1\right)\)
https://hoc24.vn/hoi-dap/question/405366.html
\(\sqrt{4-\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\left(4+\sqrt{15}\right)\)
\(=\sqrt{\left(4+\sqrt{15}\right)^2\left(4-\sqrt{15}\right)}\times\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(4+\sqrt{15}\right)\left(16-15\right)}\times\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\)
= 5 - 3
= 2
2.+ \(\left(2n+1\right)^2=4n^2+4n+1>4n^2+4n\)
\(\Rightarrow2n+1>\sqrt{4n\left(n+1\right)}=2\sqrt{n\left(n+1\right)}\)
+ \(\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(2n+1\right)\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(A< \frac{1}{2}\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{48}}-\frac{1}{\sqrt{49}}\right)\)
\(\Rightarrow A< \frac{1}{2}\)
1. + \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\left(n+1\right)-n}{\left(n+1\right)\sqrt{n}}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(n+1\right)\sqrt{n}}\)
\(< \frac{\left(\sqrt{n+1}-\sqrt{n}\right)\cdot2\sqrt{n+1}}{\sqrt{n}\left(n+1\right)}=2\cdot\frac{n+1-\sqrt{n\left(n+1\right)}}{\left(n+1\right)\sqrt{n}}=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Do đó : \(A< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
\(\Rightarrow A< 2\)
Bài 2 tạm thời chưa nghĩ ra :))
\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
\(=\sqrt{\frac{3+2\sqrt{3}\sqrt{2}+2}{3-2\sqrt{3}\sqrt{2}+2}}+\sqrt{\frac{3-2\sqrt{3}\sqrt{2}+2}{3+2\sqrt{3}\sqrt{2}+2}}\)
\(=\sqrt{\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)^2}}+\sqrt{\frac{\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}+\sqrt{3}\right)^2}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)}+\frac{\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}+\sqrt{3}\right)}\)\
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)^2+\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{5+2\sqrt{6}+5-2\sqrt{6}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=10\)
\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(=\sqrt{3}+2+\sqrt{2}-\sqrt{2}-3\)
\(=\sqrt{3}-1\)
Đặt B là tên biểu thức
Với mọi n thuộc N*, ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\) (*)
Áp dụng (*), ta được:
\(B< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)
\(=2\left(1-\frac{1}{\sqrt{2013}}\right)=2-\frac{1}{\sqrt{2013}}< 2\)
Bài 1:
Xét tử số:
\(\sqrt{14+6\sqrt{5}}-\sqrt{14-6\sqrt{5}}=\sqrt{3^2+5+2.3\sqrt{5}}-\sqrt{3^2+5-2.3\sqrt{5}}\)
\(=\sqrt{(3+\sqrt{5})^2}-\sqrt{(3-\sqrt{5})^2}=3+\sqrt{5}-(3-\sqrt{5})=2\sqrt{5}\)
Xét mẫu số:
\(\sqrt{(\sqrt{5}+1)\sqrt{6-2\sqrt{5}}}=\sqrt{(\sqrt{5}+1)\sqrt{5+1-2\sqrt{5}}}=\sqrt{(\sqrt{5}+1)\sqrt{(\sqrt{5}-1)^2}}\)
\(=\sqrt{(\sqrt{5}+1)(\sqrt{5}-1)}=\sqrt{4}=2\)
Do đó: $A=\frac{2\sqrt{5}}{2}=\sqrt{5}$
\(\frac{\left(5+\sqrt{24}\right)\left(49-20\sqrt{6}\right).\sqrt{5-2\sqrt{6}}}{9\sqrt{30}-11\sqrt{2}}=\frac{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)^2.\sqrt{5-2\sqrt{6}}}{9\sqrt{30}-11\sqrt{2}}\)
\(=\frac{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)\left(5-2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{30}-11\sqrt{2}}\)
\(=\frac{\left(25-24\right)\left(\sqrt{3}-\sqrt{2}\right)^2.\left(\sqrt{3}-\sqrt{2}\right)}{9\sqrt{30}-11\sqrt{2}}\)\(=\frac{\left(\sqrt{3}-\sqrt{2}\right)^3}{9\sqrt{30}-11\sqrt{2}}\)
Đến đây k biết làm