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a)
\(3+2\sqrt{2}=2+2\sqrt{2}+1=\left(\sqrt{2}^2\right)+2\times\sqrt{2}\times1=\left(\sqrt{2}+1\right)^2\)
mấy câu còn lại tương tự
a/ 3 + 2\(\sqrt{2}\) = 2 + 2\(\sqrt{2}\) + 1 = \(\sqrt{2}^2\) + 2\(\sqrt{2}\) + 12 = ( \(\sqrt{2}\) + 1 )2
b/ 3 - \(\sqrt{8}\) = 2 - \(\sqrt{4.2}\) + 1 = 2 - 2\(\sqrt{2}\) + 1 = \(\sqrt{2}^2\) - 2\(\sqrt{2}\) + 12
= ( \(\sqrt{2}\) - 1 )2
c/ 9 + 4\(\sqrt{5}\) = 4 + 2.2\(\sqrt{5}\) + 5 = 22 + 2.2\(\sqrt{5}\) + \(\sqrt{5}\)2
= ( 2 + \(\sqrt{5}\) )2
d/ 23 - 8\(\sqrt{7}\) = 16 - 2.4.\(\sqrt{7}\) + 7 = 42 - 2.4.\(\sqrt{7}\) + \(\sqrt{7}^2\)
= ( 4 - \(\sqrt{7}\) )2
Bài 3:
a) Ta có: \(4+2\sqrt{3}\)
\(=3+2\cdot\sqrt{3}\cdot1+1\)
\(=\left(\sqrt{3}+1\right)^2\)
b) Ta có: \(7+4\sqrt{3}\)
\(=4+2\cdot2\cdot\sqrt{3}+3\)
\(=\left(2+\sqrt{3}\right)^2\)
c) Ta có: \(9+4\sqrt{5}\)
\(=5+2\cdot\sqrt{5}\cdot2+4\)
\(=\left(\sqrt{5}+2\right)^2\)
d) Ta có: \(31+10\sqrt{6}\)
\(=25+2\cdot5\cdot\sqrt{6}+6\)
\(=\left(5+\sqrt{6}\right)^2\)
e) Ta có: \(13+4\sqrt{3}\)
\(=12+2\cdot2\sqrt{3}\cdot1+1\)
\(=\left(2\sqrt{3}+1\right)^2\)
g) Ta có: \(21+12\sqrt{3}\)
\(=12+2\cdot2\sqrt{3}\cdot3+9\)
\(=\left(2\sqrt{3}+3\right)^2\)
h) Ta có: \(29+12\sqrt{5}\)
\(=20+2\cdot2\sqrt{5}\cdot3+3\)
\(=\left(2\sqrt{5}+3\right)^2\)
i) Ta có: \(49+8\sqrt{3}\)
\(=48+2\cdot4\sqrt{3}\cdot1\)
\(=\left(4\sqrt{3}+1\right)^2\)
k) Sửa đề: \(14-6\sqrt{5}\)
Ta có: \(14-6\sqrt{5}\)
\(=9-2\cdot3\cdot\sqrt{5}+5\)
\(=\left(3-\sqrt{5}\right)^2\)
l) Ta có: \(23-8\sqrt{7}\)
\(=16-2\cdot4\cdot\sqrt{7}+7\)
\(=\left(4-\sqrt{7}\right)^2\)
m) Ta có: \(15-4\sqrt{11}\)
\(=11-2\cdot\sqrt{11}\cdot2+4\)
\(=\left(\sqrt{11}-2\right)^2\)
n) Sửa đề: \(28-10\sqrt{3}\)
Ta có: \(28-10\sqrt{3}\)
\(=25-2\cdot5\cdot\sqrt{3}+3\)
\(=\left(5-\sqrt{3}\right)^2\)
o) Ta có: \(17-12\sqrt{2}\)
\(=9-2\cdot3\cdot2\sqrt{2}+8\)
\(=\left(3-2\sqrt{2}\right)^2\)
p) Ta có: \(43-30\sqrt{2}\)
\(=25-2\cdot5\cdot3\sqrt{2}+18\)
\(=\left(5-3\sqrt{2}\right)^2\)
q) Ta có: \(51-10\sqrt{2}\)
\(=50-2\cdot5\sqrt{2}\cdot1\)
\(=\left(5\sqrt{2}-1\right)^2\)
r) Ta có: \(49-12\sqrt{5}\)
\(=45-2\cdot3\sqrt{5}\cdot2+4\)
\(=\left(3\sqrt{5}-2\right)^2\)
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)
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)\(\sqrt{23+8\sqrt{7}}=\sqrt{16+2.4.\sqrt{7}+7}=\sqrt{\left(4+\sqrt{7}\right)^2}=4+\sqrt{7}\)
a) \(9+4\sqrt{5}=\left(\sqrt{5}\right)^2+2.\sqrt{5}.2+2^2=\left(\sqrt{5}+2\right)^2\)
b) \(23-8\sqrt{7}=4^2-2.4.\sqrt{7}+\left(\sqrt{7}\right)^2=\left(4-\sqrt{7}\right)^2\)
c) \(4-2\sqrt{3}=\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2=\left(\sqrt{3}-1\right)^2\)
d) \(11+6\sqrt{2}=3^2+2.3.\sqrt{2}+\left(\sqrt{2}\right)^2=\left(3+\sqrt{2}\right)^2\)
a) \(9+4\sqrt{5}=\left(\sqrt{5}+2\right)^2\)
b) \(23-8\sqrt{7}=\left(4-\sqrt{7}\right)^2\)
c) \(4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)
d) \(11+6\sqrt{2}=\left(3+\sqrt{2}\right)^2\)