Chứng minh đẳng thức \(\sqrt{5-2\sqrt{6}}-\sqrt{5+\sqrt{6}}=-2\sqrt{2}\)
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Những câu hỏi liên quan
6 tháng 11 2016
Vế trái : \(\frac{1}{\sqrt{6}+\sqrt{7}}-\sqrt{7}=\frac{\sqrt{7}-\sqrt{6}}{\left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}-\sqrt{7}=\sqrt{7}-\sqrt{6}-\sqrt{7}=-\sqrt{6}\)
Vế phải : \(\sqrt{\left(\sqrt{5}-\sqrt{6}\right)^2}-2\sqrt{6}+\sqrt{5}=\left(\sqrt{6}-\sqrt{5}\right)-2\sqrt{6}+\sqrt{5}=-\sqrt{6}\)
Vậy ..................................
AT
7 tháng 7 2021
\(\sqrt{4x+8}+3\sqrt{x+2}=3+\dfrac{4}{5}\sqrt{25x+50}\left(x\ge-2\right)\)
\(\Rightarrow2\sqrt{x+2}+3\sqrt{x+2}-4\sqrt{x+2}=3\Rightarrow\sqrt{x+2}=3\Rightarrow x=7\)
\(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{\dfrac{4+2\sqrt{3}}{2}}+\sqrt{\dfrac{4-2\sqrt{3}}{2}}\)
\(=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}+\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}+\dfrac{\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
18 tháng 9 2016
Xét tử \(2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=2\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}\)
\(=2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}=2\sqrt{3+\sqrt{5-\left(2\sqrt{3}+1\right)}}\)
\(=2\sqrt{3+\sqrt{4-2\sqrt{3}}}=2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}=2\sqrt{3+\sqrt{3}-1}\)
\(=2\sqrt{2+\sqrt{3}}=\frac{2\sqrt{4+2\sqrt{3}}}{\sqrt{2}}=\frac{2\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}=\frac{2\left(\sqrt{3}+1\right)}{\sqrt{2}}=\sqrt{6}+\sqrt{2}\)
Suy ra VT = VP = 1
17 tháng 12 2023
\(\left(4-\sqrt{7}\right)^2=4^2-2\cdot4\cdot\sqrt{7}+7\)
\(=16-8\sqrt{7}+7=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)
\(=\left|\sqrt{5}-2\right|-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=\dfrac{\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\dfrac{3-1}{2-1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\dfrac{6\sqrt{6}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{1}{2}\sqrt{6}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{1}{2}-2=-\dfrac{3}{2}=-1,5\)
VH
15 tháng 5 2018
a)\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=1\)\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=1\)
\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}=1\)
\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=1\)
\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=1\)
\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
\(\Leftrightarrow\sqrt{1}=1\) (đpcm)
5 tháng 8 2018
a) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{1+2\sqrt{5}+\left(\sqrt{5}\right)^2}+\sqrt{1-2\sqrt{5}+\left(\sqrt{5}\right)^2}\)\(=\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(1-\sqrt{5}\right)^2}=1+\sqrt{5}-\left(1-\sqrt{5}\right)=1+\sqrt{5}-1+\sqrt{5}=2\sqrt{5}\)
KT
5 tháng 8 2018
a) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
b) \(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}\)
\(=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
\(=2\sqrt{2}+\sqrt{5}+2\sqrt{2}-\sqrt{5}=4\sqrt{2}\)
c) \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
\(=2\sqrt{2\sqrt{3}}-10\sqrt{2\sqrt{3}}+8\sqrt{2\sqrt{3}}=0\)
LT
1
6 tháng 8 2015
Biến đổi vế trái ta có:
\(\sqrt{3+\sqrt{5}-\sqrt{13+4\sqrt{3}}}=\sqrt{3+\sqrt{5}-\sqrt{\left(2\sqrt{3}+1\right)^2}}=\sqrt{3+\sqrt{5}-2\sqrt{3}-1}=\sqrt{2+\sqrt{5}-2\sqrt{3}}\)
Đề sai
8 tháng 10 2019
a) \(A=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\sqrt{2}\)
Biến đổi vế trái :
VT = \(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{\sqrt{2}\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{\sqrt{2}\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}\)
\(=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{4+2\sqrt{3}}}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{4-2\sqrt{3}}}=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\left|\sqrt{3}+1\right|}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\left|\sqrt{3}-1\right|}\)
\(=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{3}+1}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{3}+1}=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{\sqrt{3}+3}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{3-\sqrt{3}}=\frac{\sqrt{2}\left(2+\sqrt{3}\right)\left(\sqrt{3}-3\right)+\sqrt{2}\left(2-\sqrt{3}\right)\left(\sqrt{3}+3\right)}{\left(\sqrt{3}+3\right)\left(3-\sqrt{3}\right)}\)
\(=\frac{\sqrt{2}\left(6-2\sqrt{3}+3\sqrt{3}-3+6+2\sqrt{3}-3\sqrt{3}-3\right)}{9-3}=\frac{6\sqrt{2}}{6}=\sqrt{2}=VP\left(đpcm\right)\)
b) \(B=\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=8\)
Biến đổi vế trái :
VT = \(\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=\sqrt{5+\sqrt{21}}\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5+\sqrt{21}}\sqrt{5-\sqrt{21}}\)
\(=\sqrt{2}\sqrt{5+\sqrt{21}}\left(\sqrt{7}-\sqrt{3}\right)\sqrt{25-21}=\sqrt{10+2\sqrt{21}}\left(\sqrt{7}-\sqrt{3}\right)\sqrt{4}=\left|\sqrt{7}+\sqrt{3}\right|\left(\sqrt{7}-\sqrt{3}\right)2\)
\(=\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)2=\left(7-3\right)2=4.2=8=VP\left(đpcm\right)\)
Sửa đề: \(\sqrt{5-2\sqrt{6}}-\sqrt{5+2\sqrt{6}}\)
Ta có: \(\sqrt{5-2\sqrt{6}}-\sqrt{5+2\sqrt{6}}\)
\(=\sqrt{2-2\cdot\sqrt{2}\cdot\sqrt{3}+3}-\sqrt{2+2\cdot\sqrt{2}\cdot\sqrt{3}+3}\)
\(=\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}\)
\(=\left|\sqrt{2}-\sqrt{3}\right|-\left|\sqrt{2}+\sqrt{3}\right|\)
\(=\sqrt{3}-\sqrt{2}-\left(\sqrt{2}+\sqrt{3}\right)\)
\(=\sqrt{3}-\sqrt{2}-\sqrt{2}-\sqrt{3}\)
\(=-2\sqrt{2}\)(đpcm)