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Khách
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NH
4
10 tháng 9 2016
\(M>0\Leftrightarrow M^2=\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)^2=4+\sqrt{7}+4-\sqrt{7}-2\sqrt{4+\sqrt{7}}.\sqrt{4-\sqrt{7}}..\)
\(M^2=8-2.\sqrt{16-7}=8-6=3\)
\(M=\sqrt{3}.\)
29 tháng 9 2020
1) \(\left(\sqrt{6}-\sqrt{8}\right)\left(\sqrt{6}+\sqrt{8}\right)\)
\(=\left(\sqrt{6}\right)^2-\left(\sqrt{8}\right)^2\)
\(=6-8=-2\)
2) \(\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=3^2-\left(\sqrt{5}\right)^2\)
\(=9-5=4\)
29 tháng 9 2020
3) \(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
\(=\sqrt{4-4\sqrt{3}+3}+\sqrt{4+4\sqrt{3}+3}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
4) Xét ta thấy: \(2\sqrt{3}=\sqrt{12}< \sqrt{16}=4\)
=> \(2\sqrt{3}-4< 0\) => vô lý không tm đk căn
12 tháng 10 2022
a: \(=4+\sqrt{11}+\dfrac{3}{2}-\dfrac{1}{2}\sqrt{7}-4-2\sqrt{7}-\dfrac{1}{2}\sqrt{7}+\dfrac{5}{2}\)
\(=4+\sqrt{11}-3\sqrt{7}\)
b: \(VT=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y+2x+2y}{2\left(x-y\right)}\)
\(=\dfrac{2x+4\sqrt{xy}+2y}{2\left(x-y\right)}=\dfrac{x+2\sqrt{xy}+y}{x-y}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
22 tháng 6 2019
\(\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}\right)^2-2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}-1\right)^2}=|\sqrt{2}-1|=\sqrt{2}-1\)
Tương tự \(\sqrt{4-2\sqrt{3}}=\sqrt{3}-1\); \(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}\)
\(\Rightarrow BTT=\sqrt{2}-1+\sqrt{3}-1+2-\sqrt{3}=\sqrt{2}\)
TP
23 tháng 6 2019
\(\sqrt{3-2\sqrt{2}}+\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{2-2\sqrt{2}+1}+\sqrt{3-2\sqrt{3}+1}-\sqrt{4-4\sqrt{3}+3}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\sqrt{2}-1+\sqrt{3}-1-2+\sqrt{3}\)
\(=2\sqrt{3}+\sqrt{2}-4\)
3 tháng 8 2018
\(1,\sqrt{\left(-0,3\right)^2}=\sqrt{0,09}=0,3\)
\(2,-\frac{1}{2}\sqrt{\left(0,3\right)^2}=-\frac{1}{2}.0,3=-0,15\)
\(3,\sqrt{a^{10}}=\sqrt{\left(a^5\right)^2}=a^5\left(a\ge0\right)\)
\(4,\sqrt{\left(2-x\right)^2}=\left|2-x\right|=2-x\left(x\le2\right)\)
\(5,\sqrt{x^2+2x+1}=\sqrt{\left(x+1\right)^2}=\left|x+1\right|\)
\(6,\sqrt{\left(1-\sqrt{2}\right)^2}=\left|1-\sqrt{2}\right|=\sqrt{2}-1\)(Vì \(1< \sqrt{2}\))
\(7,\sqrt{11+6\sqrt{2}}=\sqrt{9+6\sqrt{2}+2}=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)
\(8,\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\left(\sqrt{7}-1\right)-\left(\sqrt{7}+1\right)\)
\(=-2\)
\(9,\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}+\sqrt{5-2\sqrt{5}+1}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+1+\sqrt{5}-1\)
\(=2\sqrt{5}\)
7 tháng 10 2022
1: \(\left(\sqrt{3}+\sqrt{7}\right)^2=10+2\sqrt{21}\)
\(\left(2+\sqrt{6}\right)^2=10+4\sqrt{6}\)
mà 2 căn 21<4 căn 6
nên căn 3+căn 7<2+căn 6
2: \(\sqrt{7}-\sqrt{5}=\dfrac{2}{\sqrt{7}+\sqrt{5}}\)
\(\sqrt{6}-2=\dfrac{2}{\sqrt{6}+2}\)
mà \(\sqrt{7}+\sqrt{5}>\sqrt{6}+2\)
nên \(\sqrt{7}-\sqrt{5}< \sqrt{6}-2\)
3: \(\sqrt{11}-\sqrt{7}=\dfrac{4}{\sqrt{11}+\sqrt{7}}\)
\(\sqrt{7}-\sqrt{3}=\dfrac{4}{\sqrt{7}+\sqrt{3}}\)
mà căn 11>căn 3
nên \(\sqrt{11}-\sqrt{7}< \sqrt{7}-\sqrt{3}\)
NH
4
NN
17 tháng 9 2020
a) \(\sqrt{11+4\sqrt{7}}-\sqrt{11-4\sqrt{7}}\)
\(=\sqrt{7+4\sqrt{7}+4}-\sqrt{7-4\sqrt{7}+4}\)
\(=\sqrt{\left(\sqrt{7}+2\right)^2}-\sqrt{\left(\sqrt{7}-2\right)^2}\)
\(=\left|\sqrt{7}+2\right|-\left|\sqrt{7}-2\right|\)
\(=\sqrt{7}+2-\sqrt{7}+2=4\)
17 tháng 9 2020
a) \(\sqrt{11+4\sqrt{7}}-\sqrt{11-4\sqrt{7}}=\sqrt{\left(2+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-2\right)^2}=2+\sqrt{7}-\sqrt{7}+2=4\)
b) \(A=\sqrt{11-4\sqrt{6}}-\sqrt{11+4\sqrt{6}}\)
\(\Rightarrow A^2=11-4\sqrt{6}-2\sqrt{\left(11-4\sqrt{6}\right)\left(11+4\sqrt{6}\right)}+11+4\sqrt{6}\)
\(A^2=22-2\sqrt{121-96}\)
\(A^2=22-2\sqrt{25}=22-2.5=12\)
\(\Rightarrow A=-\sqrt{12}\)(Chú ý \(A< 0\))
2 tháng 7 2019
Lời giải :
a) \(\sqrt{\left(0,1-\sqrt{0,1}\right)^2}\)
\(=0,1-\sqrt{0,1}\)
b) \(\sqrt{4-2\sqrt{3}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
c) \(\sqrt{3+2\sqrt{2}}=\sqrt{2+2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
d) \(\sqrt{9-4\sqrt{5}}=\sqrt{5-4\sqrt{5}+4}=\sqrt{\left(\sqrt{5}-2\right)^2}=\sqrt{5}-2\)
e) \(\sqrt{16-6\sqrt{7}}=\sqrt{9-2\cdot3\cdot\sqrt{7}+7}=\sqrt{\left(3-\sqrt{7}\right)^2}=3-\sqrt{7}\)
\(=\sqrt{7}-\sqrt{4-2.2\sqrt{7}+7}\)
\(=\sqrt{7}-\sqrt{\left(2-\sqrt{7}\right)^2}\)
\(=\sqrt{7}-\sqrt{7}+2=2\)
\(\sqrt{7}-\sqrt{11-4\sqrt{7}}\)
\(=\sqrt{7}-\sqrt{\sqrt{7}^2-2\cdot\sqrt{7}\cdot2+2^2}\)
\(=\sqrt{7}-\sqrt{\left(\sqrt{7}-2\right)^2}\)
\(=\sqrt{7}-\left|\sqrt{7}-2\right|\)
\(=\sqrt{7}-\sqrt{7}+2=2\)