Rút gọn:
\(C=\sqrt{8+2\sqrt{10+2\sqrt{5}}}\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
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TD
1
26 tháng 8 2023
2:
ĐKXĐ: x>=3
\(\Leftrightarrow\sqrt{x-3+2\cdot\sqrt{x-3}\cdot\sqrt{3}+3}+\sqrt{x-3-2\cdot\sqrt{x-3}\cdot\sqrt{3}+3}=2\sqrt{3}\)
=>\(\left|\sqrt{x-3}+\sqrt{3}\right|+\left|\sqrt{x-3}-\sqrt{3}\right|=2\sqrt{3}\)
\(\Leftrightarrow\sqrt{x-3}+\sqrt{3}+\left|\sqrt{x-3}-\sqrt{3}\right|=2\sqrt{3}\)
\(\Leftrightarrow\sqrt{x-3}+\left|\sqrt{x-3}-\sqrt{3}\right|=\sqrt{3}\)(1)
TH1: x>=6
(1) trở thành \(\sqrt{x-3}+\sqrt{x-3}-\sqrt{3}=\sqrt{3}\)
=>\(2\sqrt{x-3}=2\sqrt{3}\)
=>x-3=3
=>x=6(nhận)
TH2: 3<=x<6
Phương trình (1) sẽ là;
\(\sqrt{x-3}+\sqrt{3}-\sqrt{x-3}=\sqrt{3}\)
=>\(\sqrt{3}=\sqrt{3}\)(luôn đúng)
1:
\(A^2=8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{8^2-\left(2\sqrt{10+2\sqrt{5}}\right)^2}\)
\(=16+2\cdot\sqrt{64-4\cdot\left(10+2\sqrt{5}\right)}\)
\(=16+2\cdot\sqrt{24-8\sqrt{5}}\)
\(=16+2\cdot\sqrt{20-2\cdot2\sqrt{5}\cdot2+4}\)
\(=16+2\cdot\sqrt{\left(2\sqrt{5}-2\right)^2}\)
\(=16+2\cdot\left(2\sqrt{5}-2\right)=12+4\sqrt{5}\)
\(=10+2\cdot\sqrt{10}\cdot\sqrt{2}+2\)
\(=\left(\sqrt{10}+\sqrt{2}\right)^2\)
=>\(A=\sqrt{10}+\sqrt{2}\)
TN
2
14 tháng 7 2019
\(\frac{5}{\sqrt{2}-7}-\frac{4}{3\sqrt{2}+5}-\frac{7}{4-5\sqrt{2}}\)
LN
0
MT
1
8 tháng 8 2015
\(A^2=8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}+2.\sqrt{\left(8+2\sqrt{10+2\sqrt{5}}\right)\left(8-2\sqrt{10+2\sqrt{5}}\right)}\)
\(A^2=16+2.\sqrt{8^2-\left(2\sqrt{10+2\sqrt{5}}\right)^2}=16+2.\sqrt{24-8\sqrt{5}}=16+4.\sqrt{6-2\sqrt{5}}\)
\(A^2=16+4.\sqrt{\left(\sqrt{5}-1\right)^2}=16+4.\left(\sqrt{5}-1\right)=12+4\sqrt{5}\)
=> A = \(\sqrt{12+4\sqrt{5}}=\sqrt{2}\sqrt{6+2\sqrt{5}}=\sqrt{2}.\left(\sqrt{5}+1\right)=\sqrt{10}+\sqrt{2}\)
TA
0
2
2
Y
26 tháng 6 2023
\(\dfrac{1}{\sqrt{2}+1}-\dfrac{\sqrt{8}-\sqrt{10}}{2-\sqrt{5}}\\ =\dfrac{1}{\sqrt{2}+1}-\dfrac{\sqrt{2}.\sqrt{4}-\sqrt{2}.\sqrt{5}}{2-\sqrt{5}}\\ =\dfrac{1}{\sqrt{2}+1}-\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}\\ =\dfrac{1}{\sqrt{2}+1}-\sqrt{2}\\ =\dfrac{1-\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\\ =\dfrac{1-2-\sqrt{2}}{\sqrt{2}+1}\\ =\dfrac{-\sqrt{2}-1}{\sqrt{2}+1}\\ =\dfrac{-\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\\ =-1\)
\(C=\sqrt{\left(8+2\sqrt{10+2\sqrt{5}}\right).\left(8-2\sqrt{10+2\sqrt{5}}\right)}=\sqrt{\left(8^2-\left(2\sqrt{10+2\sqrt{5}}\right)^2\right)=\sqrt{64-4\left(10+2\sqrt{5}\right)}}\)
\(C=\sqrt{64-40-8\sqrt{5}}=\sqrt{24-8\sqrt{5}}\)
\(C=\sqrt{20-2.2.2\sqrt{5}+4}=\sqrt{\left(2\sqrt{5}-2\right)^2}\)
\(C=2\sqrt{5}-2=2\left(\sqrt{5}-1\right)\)