\(A=\frac{5+7\sqrt{5}}{\sqrt{5}}+\frac{11+\sqrt{11}}{1+\sqrt{11}}\)
K
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HT
2
KT
6 tháng 8 2018
\(\frac{5+7\sqrt{5}}{\sqrt{5}}+\frac{11+\sqrt{11}}{1+\sqrt{11}}\)
\(=\frac{\sqrt{5}.\left(\sqrt{5}+7\right)}{\sqrt{5}}+\frac{\sqrt{11}.\left(\sqrt{11}+1\right)}{1+\sqrt{11}}\)
\(=\sqrt{5}+7+\sqrt{11}\)
p/s: chúc bạn học tốt
HT
6 tháng 8 2018
ohh,mik cung hiểu kha khá r.nhưng bn có thể chỉ lm rõ hơn cho mik đc k?
8 tháng 5 2019
a, \(\sqrt{2}A=\sqrt{10-2\sqrt{3.7}}+\sqrt{10+2\sqrt{3.7}}\)
\(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{7}\right)^2}\)
\(=\left|\sqrt{7}-\sqrt{3}\right|+\left|\sqrt{7}+\sqrt{3}\right|\)
\(=\sqrt{7}-\sqrt{3}+\sqrt{3}+\sqrt{7}=2\sqrt{7}\)
\(\Rightarrow A=\sqrt{14}\)
b, \(B=\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\frac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)
\(=\sqrt{5}+\frac{\sqrt{5}}{2}=\frac{3\sqrt{5}}{2}\)
c, \(C=\left(1-\sqrt{11}\right)\left(\sqrt{11}+1\right)=1-11=-10\)
d, \(D=\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}{2-3}-\frac{\sqrt{2}\left(\sqrt{2}-\sqrt{3}\right)}{2-3}\)
\(=-2-\sqrt{6}+2-\sqrt{6}=-2\sqrt{6}\)
S
31 tháng 8 2019
\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\frac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\frac{6+2\sqrt{5}}{4}=\frac{32-6-2\sqrt{5}}{4}=\frac{26-2\sqrt{5}}{4}=\frac{14-\sqrt{5}}{2}\) \(\left(\frac{9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2-\left(\frac{9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2=\left(\frac{9-2\sqrt{14}-9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)\left(\frac{9-2\sqrt{14}+9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)=\frac{-72\sqrt{14}}{\sqrt{7}-\sqrt{2}}\)
30 tháng 10 2020
\(8\sqrt{\frac{2}{5}}+7\sqrt{\frac{5}{2}}-\frac{\sqrt{33}}{\sqrt{3}}+\frac{10}{\sqrt{11}-1}\\ =8\sqrt{\frac{10}{5^2}}+7\sqrt{\frac{10}{2^2}}-\sqrt{11}+\frac{10\left(\sqrt{11}+1\right)}{\left(\sqrt{11}-1\right)\left(\sqrt{11}+1\right)}\\ =\frac{8\sqrt{10}}{5}+\frac{7\sqrt{10}}{2}-\sqrt{11}+\frac{10\left(\sqrt{11}+1\right)}{10}\\ =\frac{16\sqrt{10}+35\sqrt{10}}{10}-\sqrt{11}+\sqrt{11}+1\\ =\frac{51\sqrt{10}}{10}+1\\ =\frac{10+51\sqrt{10}}{10}\)
HM
0
\(A=\frac{\sqrt{5}\left(\sqrt{5}+7\right)}{\sqrt{5}}+\frac{\sqrt{11}\left(\sqrt{11}+1\right)}{1+\sqrt{11}}=\sqrt{5}+\sqrt{11}+7\)