rut gon bieu thuc
\(\sqrt{5+\sqrt{13+\sqrt{5+...}}}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
4 tháng 7 2015
đk: x>=0; x khác 3
a) \(P=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-3}=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)
b) \(P=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)
ta có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)
TQ
2
14 tháng 10 2017
\(\sqrt{3-2\sqrt{2}}=\sqrt{1-2\sqrt{2}+2}=\sqrt{\left(1-\sqrt{2}\right)^2}=\left|1-\sqrt{2}\right|\)
\(\sqrt{5-2\sqrt{6}}=\sqrt{2-2\sqrt{6}+3}=\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}=\left|\sqrt{2}-\sqrt{3}\right|\)
Mà\(1< \sqrt{2};\sqrt{2}< \sqrt{3}\)
\(\Rightarrow\sqrt{3-2\sqrt{2}}+\sqrt{5-2\sqrt{6}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}\)
\(=\sqrt{3}-1\)
14 tháng 10 2017
ta có: \(\sqrt{3-2\sqrt{2}}+\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}.\)
\(\sqrt{2}-1+\sqrt{3}-\sqrt{2}=\sqrt{3}-1\)
28 tháng 8 2017
nx \(\frac{1}{\sqrt{n}+\sqrt{n+4}}\) =\(\frac{\sqrt{n+4}-\sqrt{n}}{\left(\sqrt{n+4}+\sqrt{n}\right)\left(\sqrt{n+4}-\sqrt{n}\right)}=\frac{\sqrt{n+4}-\sqrt{n}}{n+4-n}=\frac{1}{4}.\left(\sqrt{n+4}-\sqrt{n}\right)\)
ap dung ta co \(=\frac{1}{4}\left(-1+\sqrt{5}-\sqrt{5}+\sqrt{9}+...-\sqrt{2009}+\sqrt{2013}\right)\)
=\(\frac{1}{4}\left(\sqrt{2013}-1\right)\)
Q
4 tháng 6 2017
a) \(\sqrt{15+2\sqrt{5}-\sqrt{21-4\sqrt{5}}}\)
\(=\sqrt{15+2\sqrt{5}-\sqrt{\left(1-2\sqrt{5}\right)^2}}\)
\(=\sqrt{15+2\sqrt{5}-\left(2\sqrt{5}-1\right)}\)
\(=\sqrt{15+2\sqrt{5}-\left(2\sqrt{5}-1\right)}\)
\(=\sqrt{15+2\sqrt{5}-2\sqrt{5}+1}\)
\(=\sqrt{16}\)
\(=4\)
b) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt[4]{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt[4]{5-\sqrt{3-\sqrt{\left(3-2\sqrt{5}\right)^2}}}\)
\(=\sqrt[4]{5-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)
\(=\sqrt[4]{5-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt[4]{5-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt[4]{5-\sqrt{\left(1-\sqrt{5}\right)^2}}\)
\(=\sqrt[4]{5-\left(\sqrt{5}-1\right)}\)
\(=\sqrt[4]{5-\sqrt{5}+1}\)
\(=\sqrt[4]{6-\sqrt{5}}\)
8 tháng 8 2016
1) \(\frac{\sqrt{6-2\sqrt{5}}}{2-2\sqrt{5}}=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}}{2\left(1-\sqrt{5}\right)}=\frac{\sqrt{5}-1}{2\left(1-\sqrt{5}\right)}=-\frac{1}{2}\)
2) \(\frac{\sqrt{7-4\sqrt{3}}}{1-\sqrt{3}}=\frac{\sqrt{\left(2-\sqrt{3}\right)^2}}{1-\sqrt{3}}=\frac{2-\sqrt{3}}{1-\sqrt{3}}\)
MT
9 tháng 1 2016
Điều kiện : x>=0
\(\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{7+4\sqrt{3}}-x}{\sqrt[4]{9-4\sqrt{5}}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{\left(2+\sqrt{3}\right)^2}-x}{\sqrt[4]{\left(\sqrt{5}-2\right)^2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[3]{2+\sqrt{3}}-x}{\sqrt{\sqrt{5}-2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[3]{1}-x}{\sqrt{1}+\sqrt{x}}=\sqrt{x}+\frac{1-x}{1+\sqrt{x}}=\sqrt{x}+\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{1+\sqrt{x}}\)
\(=\sqrt{x}+1-\sqrt{x}=1\)
PK
2
PN
8 tháng 10 2016
A2=x+\(\sqrt{2x-1}\)+x-\(\sqrt{2x-1}\)- 2\(\sqrt{\left(x+\sqrt{2x-1}\right)\left(x-\sqrt{2x-1}\right)}\)
A2=2x-2\(\sqrt{x^2-2x+1}\)
A2=2x-2(x-1)=1
=>A=1(vì a>0)
19 tháng 9 2020
Ta có: \(A=\sqrt{x+\sqrt{2x-1}}-\sqrt{x-\sqrt{2x-1}}\) \(\left(ĐK:x\ge\frac{1}{2}\right)\)
\(\Leftrightarrow A\sqrt{2}=\sqrt{2x+2\sqrt{2x-1}}-\sqrt{2x-2\sqrt{2x-1}}\)
\(\Leftrightarrow A\sqrt{2}=\sqrt{2x-1+2\sqrt{2x-1}+1}-\sqrt{2x-1-2\sqrt{2x-1}+1}\)
\(\Leftrightarrow A\sqrt{2}=\sqrt{\left(\sqrt{2x-1}+1\right)^2}-\sqrt{\left(\sqrt{2x-1}-1\right)^2}\)
\(\Leftrightarrow A\sqrt{2}=\sqrt{2x-1}+1-\sqrt{2x-1}+1\)
\(\Leftrightarrow A\sqrt{2}=2\)
\(\Leftrightarrow A=\sqrt{2}\)
