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28 tháng 6 2019
a. \(=\sqrt{2}.\left(\sqrt{7}+\sqrt{8}\right)\sqrt{5-\sqrt{3}\sqrt{7}}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{3-2\sqrt{3}.\sqrt{7}+7}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
Rồi nhân ra. bạn làm tiếp nhé. Tuy nhiên minh nghĩ bạn bị nhầm đề. là \(\sqrt{6}\) chứ không phải căn 16
b. \(=\frac{5\left(\sqrt{21}+1\right)}{21-16}+\frac{\sqrt{3}.\sqrt{7}\left(\sqrt{3}-\sqrt{7}\right)}{-\left(\sqrt{3}-\sqrt{7}\right)}\)
\(=\sqrt{21}+4-\sqrt{21}=4\)
HH
4 tháng 12 2020
Làm luôn nhé
\(2B=21.2\left[\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)\right]^2-2.15\sqrt{15}\)
\(2B=21\left(\sqrt{3}+1+\sqrt{5}-1\right)^2-6\left(\sqrt{3}-1+\sqrt{5}-1\right)^2-30\sqrt{15}\)
\(2B=21\left(\sqrt{3}+\sqrt{5}\right)^2-6\left(\sqrt{3}+\sqrt{5}\right)^2-30\sqrt{15}\)
\(2B=15\left(\sqrt{3}+\sqrt{5}\right)^2-30\sqrt{15}\)
\(2B=15\left(8+2\sqrt{15}\right)-30\sqrt{15}\)
\(2B=120+30\sqrt{15}-30\sqrt{5}\)
\(2B=120\)
\(B=60\)
9 tháng 7 2017
\(\text{a)}\)\(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)
\(\Leftrightarrow5\sqrt{10}+10-\sqrt{250}\)
\(\Leftrightarrow5\sqrt{10}+10-5\sqrt{10}\)
\(\Leftrightarrow10\)
\(\text{b)}\)\(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
\(\Leftrightarrow4\sqrt{21}-2\sqrt{21}-7+2\sqrt{21}\)
\(\Leftrightarrow4\sqrt{21}-7\)
KT
26 tháng 7 2018
\(A=\sqrt{\frac{5+\sqrt{21}}{5-\sqrt{21}}}+\sqrt{\frac{5-\sqrt{21}}{5+\sqrt{21}}}\)
\(=\sqrt{\frac{\left(5+\sqrt{21}\right)^2}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}}+\sqrt{\frac{\left(5-\sqrt{21}\right)^2}{\left(5-\sqrt{21}\right)\left(5+\sqrt{21}\right)}}\)
\(=\sqrt{\frac{\left(5+\sqrt{21}\right)^2}{4}}+\sqrt{\frac{\left(5-\sqrt{21}\right)^2}{4}}\)
\(=\frac{5+\sqrt{21}}{2}+\frac{5-\sqrt{21}}{2}=5\)
\(B=\sqrt{7+\sqrt{33}}+\sqrt{7-\sqrt{33}}\)
\(\Rightarrow\)\(\sqrt{2}B=\sqrt{14+2\sqrt{33}}+\sqrt{14-2\sqrt{33}}\)
\(=\sqrt{\left(\sqrt{11}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}\)
\(=\sqrt{11}+\sqrt{3}+\sqrt{11}-\sqrt{3}=2\sqrt{11}\)
\(\Rightarrow\)\(B=\sqrt{22}\)
H9
HT.Phong (9A5)
CTVHS
17 tháng 7 2023
1) \(\sqrt{6+4\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{2^2+2\cdot2\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{3^2-2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=\left|2+\sqrt{2}\right|-\left|3-\sqrt{2}\right|\)
\(=2+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}-1\)
2) \(\sqrt{21-4\sqrt{5}}+\sqrt{21+4\sqrt{5}}\)
\(=\sqrt{20-4\sqrt{5}+1}+\sqrt{20+4\sqrt{5}+1}\)
\(=\sqrt{\left(2\sqrt{5}\right)^2-2\sqrt{5}\cdot2\cdot1+1^2}+\sqrt{\left(2\sqrt{5}\right)^2+2\sqrt{5}\cdot2\cdot1-1^2}\)
\(=\sqrt{\left(2\sqrt{5}-1\right)^2}+\sqrt{\left(2\sqrt{5}+1\right)^2}\)
\(=\left|2\sqrt{5}-1\right|+\left|2\sqrt{5}+1\right|\)
\(=2\sqrt{5}-1+2\sqrt{5}+1\)
\(=4\sqrt{5}\)
LH
1
CM
12 tháng 12 2017
\(\sqrt{21-8\sqrt{5}}\)\(-\sqrt{21-4\sqrt{5}}\)
\(=\sqrt{16-2.4\sqrt{5}+5}\)\(-\sqrt{20-2\sqrt{20}+1}\)
\(=\sqrt{\left(4-\sqrt{5}\right)^2}\)\(-\sqrt{\left(\sqrt{20}-1\right)}\)
\(=4-\sqrt{5}-\left(\sqrt{20}-1\right)\)
\(=4-\sqrt{5}-\sqrt{20}+1\)
\(=5-\sqrt{5}-2\sqrt{5}\)
\(=5-3\sqrt{5}\)
TN
1
26 tháng 7 2018
\(A=\sqrt{5+\sqrt{21}}+\sqrt{5-\sqrt{21}}\)
\(=\sqrt{\frac{2\left(5+\sqrt{21}\right)}{2}}+\sqrt{\frac{2\left(5-\sqrt{21}\right)}{2}}\)
\(=\sqrt{\frac{10+2\sqrt{21}}{2}}+\sqrt{\frac{10-2\sqrt{21}}{2}}\)
\(=\sqrt{\frac{3+2\sqrt{21}+7}{2}}+\sqrt{\frac{3-2\sqrt{21+7}}{2}}\)
\(=\sqrt{\frac{\left(\sqrt{3}+\sqrt{7}\right)}{2}}+\sqrt{\frac{\left(\sqrt{3}+\sqrt{7}\right)}{2}}\)
\(=\frac{\left|\sqrt{3}+\sqrt{7}\right|}{2}+\frac{\left|\sqrt{3}-\sqrt{7}\right|}{2}\)
\(=\frac{\sqrt{3}+\sqrt{7}-\sqrt{3}-\sqrt{7}}{2}\)
\(=\sqrt{\frac{6}{2}}=\sqrt{3}\)
\(\sqrt{5+\sqrt{21}}-\sqrt{5-\sqrt{21}}\\ =\dfrac{\left(\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\right)}{\sqrt{2}}\\ =\dfrac{\left(\sqrt{7+2\sqrt{7}.\sqrt{3}+3}-\sqrt{7-2\sqrt{7}.\sqrt{3}+3}\right)}{\sqrt{2}}\\ =\dfrac{\sqrt{7}+\sqrt{3}-\sqrt{7}+\sqrt{3}}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\left|\sqrt{7}+\sqrt{3}\right|-\left|\sqrt{7}-\sqrt{3}\right|\right)\)
\(=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)