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28 tháng 6 2019
a)(\(\sqrt{2006}-\sqrt{2005}\)).(\(\sqrt{2006}+\sqrt{2005}\))
=\(\sqrt{2006}^2-\sqrt{2005}^2\)
=2006-2005
=1
NN
13 tháng 5 2018
a)\(\sqrt{13-4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{12-2.2\sqrt{3}.1+1}+\sqrt{4-2.2.\sqrt{3}+3}\)
\(=\sqrt{\left(2\sqrt{3}-1\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left|2\sqrt{3}-1\right|+\left|2-\sqrt{3}\right|\)
\(=2\sqrt{3}-1+2-\sqrt{3}=\sqrt{3}+1\)
b)\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5+2\sqrt{5}.1+1}+\sqrt{5-2\sqrt{5}.1+1}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\left(\sqrt{5}+1\right)+\left(\sqrt{5}-1\right)=2\sqrt{5}\)
c)\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{3+2\sqrt{3}.1+1}-\sqrt{3-2\sqrt{3}.1+1}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3}+1\right)-\left(\sqrt{3}-1\right)=2\)
d)\(\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{4+2.2\sqrt{3}+3}+\sqrt{4-2.2.\sqrt{3}+3}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left(2+\sqrt{3}\right)+\left(2-\sqrt{3}\right)=4\)
e)\(\sqrt{9+4\sqrt{5}}=\sqrt{5+2.\sqrt{5}.2+4}=\sqrt{\left(\sqrt{5}+2\right)^2}=\sqrt{5}+2\)
f)\(\sqrt{23+8\sqrt{7}}=\sqrt{16+2.4.\sqrt{7}+7}=\sqrt{\left(4+\sqrt{7}\right)^2}=4+\sqrt{7}\)
VT
1
8 tháng 8 2020
\(\sqrt{2}D=\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{8}\right)\sqrt{2}\)
\(\sqrt{2}D=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}+4\)
\(\sqrt{2}D=\sqrt{7}-1-\sqrt{7}-1+4\)
11 tháng 7 2020
Câu 8:
a)
Ta có: \(VT=\sqrt{4-2\sqrt{3}}-\sqrt{3}\)
\(=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}-\sqrt{3}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)
\(=\left|\sqrt{3}-1\right|-\sqrt{3}\)(1)
Ta có: 3>1
\(\Leftrightarrow\sqrt{3}>\sqrt{1}\)
\(\Leftrightarrow\sqrt{3}>1\)
\(\Leftrightarrow\sqrt{3}-1>0\)
\(\Leftrightarrow\left|\sqrt{3}-1\right|=\sqrt{3}-1\)(2)
Từ (1) và (2) suy ra \(VT=\sqrt{3}-1-\sqrt{3}=-1=VP\)(đpcm)
b) Ta có: \(VP=\left(\sqrt{5}+2\right)^2\)
\(=\left(\sqrt{5}\right)^2+2\cdot\sqrt{5}\cdot2+2^2\)
\(=5+4\sqrt{5}+4\)
\(=9+4\sqrt{5}=VT\)(đpcm)
c) Ta có: \(VT=\sqrt{9+4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{4+2\cdot2\cdot\sqrt{5}+5}-\sqrt{5}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{5}\)
\(=\left|2+\sqrt{5}\right|-\sqrt{5}\)
\(=2+\sqrt{5}-\sqrt{5}=2=VP\)(đpcm)
d) Ta có: \(VT=\sqrt{23+8\sqrt{7}}-\sqrt{7}\)
\(=\sqrt{16+2\cdot4\cdot\sqrt{7}+7}-\sqrt{7}\)
\(=\sqrt{\left(4+\sqrt{7}\right)^2}-\sqrt{7}\)
\(=\left|4+\sqrt{7}\right|-\sqrt{7}\)
\(=4+\sqrt{7}-\sqrt{7}\)
\(=4=VP\)(đpcm)
\(D=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{8}\)
=> \(\sqrt{2}.D=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}+\sqrt{8}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{8}\)
\(=\sqrt{7}-1-\sqrt{7}-1+\sqrt{8}\)
\(=2\sqrt{2}-2=\sqrt{2}.\left(2-\sqrt{2}\right)\)
=> \(D=2-\sqrt{2}\)