Tính
\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
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TH
1
9 tháng 7 2021
\(=>M^2=4-\sqrt{10-2\sqrt{5}}+2\sqrt{\left(4-\sqrt{10-2\sqrt{5}}\right)\left(4+\sqrt{10-2\sqrt{5}}\right)}\)
\(+4+\sqrt{10-2\sqrt{5}}\)
\(M^2=8+2\)\(\sqrt{16-\left(\sqrt{10-2\sqrt{5}}\right)^2}\)\(=8+2\sqrt{16-10+2\sqrt{5}}\)
\(=>M^2=8+2\sqrt{6+2\sqrt{5}}=8+2\sqrt{\left(\sqrt{5}+1\right)^2}=8+2\sqrt{5}+2\)
\(=10+2\sqrt{5}\)
\(=>M=\sqrt{10+2\sqrt{5}}\)
LL
1
AH
Akai Haruma
Giáo viên
14 tháng 10 2021
Lời giải:
Gọi biểu thức trên là $A$
\(A^2=8+2\sqrt{(4+\sqrt{10+2\sqrt{5}})(4-\sqrt{10+2\sqrt{5}})}\)
\(=8+2\sqrt{4^2-(10+2\sqrt{5})}=8+2\sqrt{6-2\sqrt{5}}\)
\(=8+2\sqrt{(\sqrt{5}-1)^2}=8+2|\sqrt{5}-1|=6+2\sqrt{5}=(\sqrt{5}+1)^2\)
$\Rightarrow A=\sqrt{5}+1$ (do $A>0$)
NV
1
HM
1
6 tháng 8 2020
Ta có:\(\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)^2\)=
=\(4+\sqrt{10+2\sqrt{5}}+2.\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right).\left(4-\sqrt{10+2\sqrt{5}}\right)}+\)\(4-\sqrt{10+2\sqrt{5}}\)
=\(8\)\(+2.\sqrt{16-10-2\sqrt{5}}\)
=\(8+2\sqrt{6-2\sqrt{5}}\)
=\(8+2.\sqrt{5-2\sqrt{5}+1}\)
=\(8+2.\sqrt{\left(\sqrt{5}-1\right)^2}\)
=\(8+2.\left(\sqrt{5}-1\right)\)
=\(8+2\sqrt{5}-2\)
=\(6+2\sqrt{5}\)
=\(\left(\sqrt{5}+1\right)^2\)
\(\Rightarrow A=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)
6 tháng 7 2021
1.\(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3=2\sqrt{2}+6+3\sqrt{2}+1-\left(2\sqrt{2}-6+3\sqrt{2}-1\right)=14\)
2.\(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
\(=\sqrt{\dfrac{1}{2}\left(8-2\sqrt{3.}\sqrt{5}\right)}+\sqrt{\dfrac{1}{2}\left(8+2.\sqrt{3}.\sqrt{5}\right)}-\sqrt{2}\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\dfrac{1}{2}\left(\sqrt{3}-\sqrt{5}\right)^2}+\sqrt{\dfrac{1}{2}\left(\sqrt{3}+\sqrt{5}\right)^2}-\sqrt{2}\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\dfrac{\sqrt{2}}{2}\left|\sqrt{3}-\sqrt{5}\right|+\dfrac{\sqrt{2}}{2}\left(\sqrt{3}+\sqrt{5}\right)-\sqrt{2}\left|\sqrt{5}-1\right|\)
\(=\dfrac{\sqrt{2}}{2}\left(\sqrt{5}-\sqrt{3}\right)+\dfrac{\sqrt{2}}{2}\left(\sqrt{3}+\sqrt{5}\right)-\sqrt{2}\left(\sqrt{5}-1\right)\)
\(=\sqrt{5}.\sqrt{2}-\sqrt{2}\left(\sqrt{5}-1\right)=\sqrt{2}\)
3.\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{1-\left(\sqrt{5}\right)^2}\)
\(=\sqrt{20}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}=2\sqrt{5}-2\left(1+\sqrt{5}\right)=-2\)
4.\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\sqrt{\dfrac{4-2\sqrt{3}}{4+2\sqrt{3}}}+\sqrt{\dfrac{4+2\sqrt{3}}{4-2\sqrt{3}}}\)\(=\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}+1\right)^2}}+\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}-1\right)^2}}\)
\(=\dfrac{\left|\sqrt{3}-1\right|}{\sqrt{3}+1}+\dfrac{\sqrt{3}+1}{\left|\sqrt{3}-1\right|}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}+\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(=\dfrac{\left(\sqrt{3}-1\right)^2+\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{8}{3-1}=4\)
6 tháng 7 2021
3: Ta có: \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)
\(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{8\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
\(=2\sqrt{5}-2\left(\sqrt{5}+1\right)\)
=-2
4) Ta có: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}\)
=4
NN
2
24 tháng 11 2018
\(A^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{16-10-2\sqrt{5}}\)
\(A^2=8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{5-2\sqrt{5}+1}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(A^2=8+2\left|\sqrt{5}-1\right|=8+2\left(\sqrt{5}-1\right)=8+2\sqrt{5}-2=6+2\sqrt{5}\)
\(\Rightarrow\)\(A=\sqrt{A^2}=\sqrt{5+2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}+1\right)^2}=\left|\sqrt{5}+1\right|=\sqrt{5}+1\)
...
HM
1
E
20 tháng 5 2017
Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}.\)
=>\(A^2=4+\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}+4-\)\(\sqrt{10+2\sqrt{5}}.\)
=\(8+2\sqrt{16-10-2\sqrt{5}}=8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\left(\sqrt{5}-1\right)\)
=\(8+2\sqrt{5}-2=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)
=>\(\sqrt{5}+1\)
HP
1
21 tháng 6 2017
\(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Leftrightarrow A^2=8+2\sqrt{4+\sqrt{10+2\sqrt{5}}}.\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Leftrightarrow A^2=8+2\sqrt{6-2\sqrt{5}}=8+2\left(\sqrt{5}-1\right)\)
\(=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)
\(\Rightarrow A=\sqrt{5}+1\)
LT
1
22 tháng 9 2020
Đặt \(D=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Leftrightarrow D^2=8+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(\Leftrightarrow D^2=8+2\sqrt{16-10-2\sqrt{5}}\)
\(\Leftrightarrow D^2=8+2\sqrt{6-2\sqrt{5}}\)
\(\Leftrightarrow D^2=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(\Leftrightarrow D^2=8+2\left(\sqrt{5}-1\right)\)
\(\Leftrightarrow D^2=6+2\sqrt{5}\)
\(\Leftrightarrow D^2=\left(\sqrt{5}+1\right)^2\)
\(\Rightarrow D=\sqrt{5}+1\)
Thay vào ta tính được: \(A=\sqrt{5}+1-\sqrt{5}=1\)
Vậy A = 1
Đặt biểu thứa là A
Bình phương 2 vế ta dc:
\(8+2\sqrt{4+\sqrt{10+2\sqrt{5}}\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(=8+2\sqrt{\left(10-\left(10+2\sqrt{5}\right)\right)}\)
\(A^2=8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=8+2\left(\sqrt{5}-1\right)\)
Do A>0 nên :
\(A=\sqrt{8+2\left(\sqrt{5}-1\right)}=6+2\sqrt{5}=\sqrt{5}+1\)
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