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TH
12 tháng 9 2018
\(A=0.5\cdot4\sqrt{3-x}-\sqrt{3-x}-2\sqrt{3}+1=\sqrt{3-x}-2\sqrt{3}+1\) (xác định khi x=<3)
a)thay \(x=2\sqrt{2}\)vào a ra có
\(\sqrt{3-2\sqrt{2}}-2\sqrt{3}+1=\sqrt{\left(\sqrt{2}-1\right)^2}-2\sqrt{3}+1\)
\(=\sqrt{2}-1+2\sqrt{3}+1=\sqrt{2}+2\sqrt{3}\)
Để A=1<=> \(\sqrt{3-x}-2\sqrt{3}+1=1\\ \Leftrightarrow\sqrt{3-x}-2\sqrt{3}+1-1=0\\ \Leftrightarrow\sqrt{3-x}-2\sqrt{3}=0\\ \Leftrightarrow3-x=12\Leftrightarrow x=-9\)
24 tháng 11 2015
\(=\sqrt{\frac{\sqrt{5}+2}{\sqrt{5}+1}}+\sqrt{\frac{\sqrt{5}-2}{\sqrt{5}+1}}-\sqrt{3-2\sqrt{3}.\sqrt{2}+2}\)
\(=\sqrt{\frac{3+\sqrt{5}}{4}}+\sqrt{\frac{7-3\sqrt{5}}{4}}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\sqrt{\frac{6+2\sqrt{5}}{8}}+\sqrt{\frac{14-2.3\sqrt{5}}{8}}-\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\sqrt{\frac{\left(\sqrt{5}+1\right)^2}{8}}+\sqrt{\frac{\left(3-\sqrt{5}\right)^2}{8}}-\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\frac{\sqrt{5}+1+3-\sqrt{5}}{2\sqrt{2}}-\sqrt{3}+\sqrt{2}\)=\(\sqrt{2}-\sqrt{3}+\sqrt{2}=-\sqrt{3}\)
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 = \(\sqrt{12+4\sqrt{5}}=\sqrt{2}\sqrt{6+2\sqrt{5}}=\sqrt{2}.\left(\sqrt{5}+1\right)=\sqrt{10}+\sqrt{2}\)
KT
21 tháng 7 2018
a) \(\sqrt{9+4\sqrt{5}}.\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+2\right)^2}.\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)\)
\(=1\)
b) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(=\left(4+\sqrt{15}\right).\left(8-2\sqrt{15}\right)\)
\(=\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right).2\)
\(=2\)
c) \(\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{6-2\sqrt{5}}.\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)
\(=\left(\sqrt{5}-1\right)^2.\left(3+\sqrt{5}\right)\)
\(=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=2\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=2.4=8\)
p/s: lần sau bạn đăng đề ghi rõ ràng ra nhé, đặc biệt biểu thức trong căn
TT
6 tháng 6 2016
\(A=\left(\sqrt{5}+2\right)\frac{5+4\sqrt{5}+4-8\sqrt{5}}{\sqrt{5}-2}\\ A=\left(\sqrt{5}+2\right)\frac{5-4\sqrt{5}+4}{\sqrt{5}-2}\\ A=\left(\sqrt{5}+2\right)\frac{\left(\sqrt{5}-2\right)^2}{\sqrt{5}-2}\\ A=\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)\\ A=1\)
NN
1
NT
0
-(5*căn bậc hai(2)*căn bậc hai(3)-17*căn bậc hai(2)-5)/(5*căn bậc hai(2))
\(\frac{2}{5-\sqrt{3}}+\frac{5}{2\sqrt{2}+3}-\sqrt{8}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{22}+\frac{5\left(2\sqrt{2}-3\right)}{-1}-\sqrt{8}\)
\(=\frac{5+\sqrt{3}}{11}-10\sqrt{2}+15-2\sqrt{2}\)
\(=\frac{5+\sqrt{3}}{11}-12\sqrt{2}+15=\frac{5+\sqrt{3}}{11}-\frac{11\left(12\sqrt{2}-15\right)}{11}\)
\(=\frac{5+\sqrt{3}-132\sqrt{2}+165}{11}=\frac{\sqrt{3}-132\sqrt{2}+170}{11}\)