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AH
Akai Haruma
Giáo viên
26 tháng 8 2023
Lời giải:
a.
\(=\sqrt{5+2.2\sqrt{5}+2^2}-\sqrt{5-2.2\sqrt{5}+2^2}\)
$=\sqrt{(\sqrt{5}+2)^2}-\sqrt{(\sqrt{5}-2)^2}$
$=|\sqrt{5}+2|-|\sqrt{5}-2|=(\sqrt{5}+2)-(\sqrt{5}-2)=4$
b.
$=\sqrt{3-2.3\sqrt{3}+3^2}+\sqrt{3+2.3.\sqrt{3}+3^2}$
$=\sqrt{(\sqrt{3}-3)^2}+\sqrt{(\sqrt{3}+3)^2}$
$=|\sqrt{3}-3|+|\sqrt{3}+3|$
$=(3-\sqrt{3})+(\sqrt{3}+3)=6$
c.
$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$
$=\sqrt{(\sqrt{2}+3)^2}-\sqrt{(\sqrt{2}-3)^2}$
$=|\sqrt{2}+3|-|\sqrt{2}-3|$
$=(\sqrt{2}+3)-(3-\sqrt{2})=2\sqrt{2}$
27 tháng 7 2023
b: =căn 10-3+4-căn 10=1
a: \(=\sqrt{11-4\sqrt{6}+\sqrt{15}}\)
12 tháng 7 2021
a) Ta có: \(\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)
\(=\sqrt{2}-1-3-\sqrt{2}\)
=-4
b) Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}\)
\(=\sqrt{3}-1-2+\sqrt{3}+4+\sqrt{3}\)
\(=3\sqrt{3}+1\)
c) Ta có: \(\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}\)
\(=\sqrt{5}-1+\sqrt{5}-2-3+\sqrt{5}\)
\(=3\sqrt{5}-6\)
d) Ta có: \(\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2\right)^6}\)
\(=\sqrt{7}-2+4-\sqrt{7}+8\)
=10
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 \(A< 0\))
27 tháng 8 2023
a: \(=\left(\sqrt{3}-2\right)\cdot\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)\)
=3-4=-1
b: \(=\sqrt{6+4\sqrt{2}}-\sqrt{11-2\sqrt{18}}\)
\(=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=2+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}-1\)
c: \(=\sqrt{\left(2\sqrt{5}-1\right)^2}+\sqrt{\left(2\sqrt{5}+1\right)^2}\)
\(=2\sqrt{5}-1+2\sqrt{5}+1\)
\(=4\sqrt{5}\)
AT
2 tháng 7 2021
e) \(\sqrt{x^2}=\left|-8\right|\Rightarrow\left|x\right|=8\Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)
e) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{2}=\sqrt{\dfrac{8-2\sqrt{7}}{2}}-\sqrt{\dfrac{8+2\sqrt{7}}{2}}+\sqrt{2}\)
\(=\sqrt{\dfrac{\left(\sqrt{7}\right)^2-2.\sqrt{7}.1+1^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}\right)^2+2.\sqrt{7}.1+1^2}{2}}+\sqrt{2}\)
\(=\sqrt{\dfrac{\left(\sqrt{7}-1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{7}+1\right)^2}{2}}+\sqrt{2}\)
\(=\dfrac{\left|\sqrt{7}-1\right|}{\sqrt{2}}-\dfrac{\left|\sqrt{7}+1\right|}{\sqrt{2}}+\sqrt{2}=\dfrac{\sqrt{7}-1}{\sqrt{2}}-\dfrac{\sqrt{7}+1}{\sqrt{2}}+\sqrt{2}\)
\(=-\dfrac{2}{\sqrt{2}}+\sqrt{2}=-\sqrt{2}+\sqrt{2}=0\)
f) \(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}+3\sqrt{2}\)
\(=\sqrt{\dfrac{12+2\sqrt{11}}{2}}-\sqrt{\dfrac{12-2\sqrt{11}}{2}}+3\sqrt{2}\)
\(=\sqrt{\dfrac{\left(\sqrt{11}\right)^2+2.\sqrt{11}.1+1^2}{2}}-\sqrt{\dfrac{\left(\sqrt{11}\right)^2-2.\sqrt{11}.1+1^2}{2}}+3\sqrt{2}\)
\(=\sqrt{\dfrac{\left(\sqrt{11}+1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{11}-1\right)^2}{2}}+3\sqrt{2}\)
\(=\dfrac{\left|\sqrt{11}+1\right|}{\sqrt{2}}-\dfrac{\left|\sqrt{11}-1\right|}{\sqrt{2}}+3\sqrt{2}=\dfrac{\sqrt{11}+1}{\sqrt{2}}-\dfrac{\sqrt{11}-1}{\sqrt{2}}+3\sqrt{2}\)
\(=\dfrac{2}{\sqrt{2}}+3\sqrt{2}=\sqrt{2}+3\sqrt{2}=4\sqrt{2}\)
25 tháng 9 2021
1) \(=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
2) \(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}\)
3) \(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\sqrt{5}-\sqrt{2}\)
5) \(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)
6) \(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\)
7) \(=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)