Thực hiện phép tính sau:
(\(\sqrt{325}-\sqrt{117}+2\sqrt{208}\)) : \(\sqrt{13}\)
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TH
1
AN
0
11 tháng 8 2020
Đặt: \(A=\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\)
=> \(A^2=\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)
=> \(A^2=2\sqrt{5}+2\sqrt{5-4}\)
=> \(A^2=2\sqrt{5}+2\)
=> \(A^2=2\left(\sqrt{5}+1\right)\)
=> \(A=\sqrt{2\left(\sqrt{5}+1\right)}\)
=> \(\frac{A}{\sqrt{\sqrt{5}+1}}=\frac{\sqrt{2\left(\sqrt{5}+1\right)}}{\sqrt{\sqrt{5}+1}}=\sqrt{2}\)
Đặt: \(B=\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\)
=> \(VT=\frac{A}{\sqrt{\sqrt{5}+1}}-B=\sqrt{2}-\left(\sqrt{2}-1\right)=\sqrt{2}-\sqrt{2}+1=1\)
VẬY KẾT QUẢ CỦA PHÉP TÍNH = 1.
14 tháng 7 2019
\(\sqrt{24+8\sqrt{5}}+\) \(\sqrt{9-4\sqrt{5}}=\) \(\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}.2+4}\) + \(\sqrt{5-2\sqrt{5}.2+4}\)
= \(\sqrt{\left(2\sqrt{5}+2\right)^2}+\) \(\sqrt{\left(\sqrt{5}-2\right)^2}\) = \(2\sqrt{5}+2+\sqrt{5}-2=3\sqrt{5}\)
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\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\) = \(\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)= \(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
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\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
= \(\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\) \(=\sqrt{\left(3\sqrt{2}+5\right)^2}=3\sqrt{2}+5\)
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J
5 tháng 8 2017
\(A=\sqrt{2+2\sqrt{\frac{3}{4}}}+\sqrt{2-2\sqrt{\frac{3}{4}}}\)
\(A=\sqrt{\left(\sqrt{\frac{3}{2}}\right)^2+2\sqrt{\frac{3}{2}.\frac{1}{2}}+\left(\sqrt{\frac{1}{2}}\right)^2}-\sqrt{\left(\sqrt{\frac{3}{2}}\right)^2-2\sqrt{\frac{3}{2}.\frac{1}{2}}+\left(\sqrt{\frac{1}{2}}\right)^2}\)
\(A=\sqrt{\left(\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}\right)^2}-\sqrt{\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}\right)^2}\)
\(A=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}-\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}\)
\(A=2\sqrt{\frac{3}{2}}=\sqrt{4.\frac{3}{2}}=\sqrt{6}\)
HP
22 tháng 5 2018
\(A=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=|2+\sqrt{3}|+|2-\sqrt{3}|\)\(=2+\sqrt{3}+2-\sqrt{3}=4\)
TT
1
22 tháng 10 2015
\(\sqrt{10^2-6^2}-\sqrt{13^2-12^2}+\sqrt{13^2}-\sqrt{12^2}\)
\(=\sqrt{100-36}-\sqrt{169-144}+\sqrt{13^2}-\sqrt{12^2}\)
\(=\sqrt{64}-\sqrt{25}+\sqrt{13^2}-\sqrt{12^2}\)
\(=\sqrt{8^2}-\sqrt{5^2}+\sqrt{13^2}-\sqrt{12^2}\)
\(=8-5+13-12=4\)
7 tháng 8 2018
\(a,=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\left(\sqrt{20}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-\sqrt{20}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)
\(=\sqrt{1}=1\)
b,c
\(\sqrt{13+4\sqrt{3}}=\sqrt{13+2\sqrt{12}}=\sqrt{12}+1=2\sqrt{3}+1\)
=>BT=\(\sqrt{5-\left(2\sqrt{3}+1\right)}+\sqrt{3+\left(2\sqrt{3}+1\right)}\)
\(=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)
c,\(=\sqrt{1+\sqrt{3+2\sqrt{3}+1}}+\sqrt{1-\sqrt{3-\left(2\sqrt{3}-1\right)}}\)
\(=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
10 tháng 6 2017
ta thấy A> 2
Xét A2 = 5 + \(\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5...........}}}}\)
( A2 - 5 )2 = 13 + A
<=> A4 - 10A2 - A + 12 = 0
<=> (A4 - 9A2 ) - ( A2 - 9 ) - (A - 3) = 0
<=> (A - 3) [(A + 3)(A+1)(A-1)-1] =0
Vì A> 2 => (A + 3)(A+1)( A-1)-1 > 0
Do đó A - 3 = 0 <=> A = 3
A=\(\dfrac{\sqrt{325}-\sqrt{117}+2\sqrt{208}}{\sqrt{13}}=\dfrac{\sqrt{13.25}-\sqrt{13.9}+2\sqrt{13.16}}{\sqrt{13}}\)
=\(\dfrac{5\sqrt{13}-3\sqrt{13}+8\sqrt{13}}{\sqrt{13}}=\dfrac{10\sqrt{13}}{\sqrt{13}}=10\)