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20 tháng 7 2017
a. \(\dfrac{3\sqrt{7}+7\sqrt{3}}{\sqrt{21}}=\dfrac{\sqrt{21}\left(\sqrt{3}+\sqrt{7}\right)}{\sqrt{21}}=\sqrt{7}+\sqrt{3}\)
b. \(\dfrac{2\sqrt{5}-4\sqrt{10}}{3\sqrt{10}}=\dfrac{\sqrt{10}\left(\sqrt{2}-4\right)}{3\sqrt{10}}=\dfrac{-4+\sqrt{2}}{3}\)
c. \(\dfrac{3-\sqrt{7}}{3+\sqrt{7}}-\dfrac{3+\sqrt{7}}{3-\sqrt{7}}=\dfrac{\left(3-\sqrt{7}\right)^2}{9-7}-\dfrac{\left(3+\sqrt{7}\right)^2}{9-7}=\dfrac{\left(3-\sqrt{7}-3-\sqrt{7}\right)\left(3-\sqrt{7}+3+\sqrt{7}\right)}{2}=\dfrac{-2\sqrt{7}.6}{2}=-6\sqrt{7}\)
L
5 tháng 12 2019
\(a)\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)
\(=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2+\left(\sqrt{7}+\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\)
\(=\frac{7+2\sqrt{35}+5+7-2\sqrt{35}+5}{7-5}\)
\(=\frac{24}{2}\)
\(=12\)
\(b)\frac{4+\sqrt{2}-\sqrt{3}-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{2}-\sqrt{3}\right)+\left(2+\sqrt{8}-\sqrt{6}\right)}{2+\sqrt{2}-\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{2}-\sqrt{3}\right)+\sqrt{2}\left(\sqrt{2}+2-\sqrt{3}\right)}{2+\sqrt{2}-\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}\right)}{2+\sqrt{2}-\sqrt{3}}\)
\(=1+\sqrt{2}\)
\(c)A=\left(\sqrt{3}+1\right)\sqrt{\frac{14-6\sqrt{3}}{5+\sqrt{3}}}\)
\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{\left(14-6\sqrt{3}\right)\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}}\)
\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{70-14\sqrt{3}-30\sqrt{3}+18}{25-3}}\)
\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{88-44\sqrt{3}}{22}}\)
\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{44\left(2-\sqrt{3}\right)}{22}}\)
\(A=\left(\sqrt{3}+1\right)\sqrt{2\left(2-\sqrt{3}\right)}\)
\(A=\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}\)
\(A=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(A=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)
\(A=3-1=2\)
P/s: nếu đề là vậy thì t ra kết quả như vậy ạ, nhưng lần sau khi đăng câu hỏi bạn nên viết rõ hơn ra nhé
HC
31 tháng 5 2017
ta có
\(\sqrt{\dfrac{7}{3}}+\sqrt{\dfrac{5}{3}}+1=\dfrac{\sqrt{7}+\sqrt{5}+\sqrt{3}}{\sqrt{3}}\)
tương tự ta có
\(\sqrt{\dfrac{3}{5}}+\sqrt{\dfrac{7}{5}}+1=\dfrac{\sqrt{3}+\sqrt{5}+\sqrt{7}}{\sqrt{5}}\)
\(\sqrt{\dfrac{3}{7}}+\sqrt{\dfrac{5}{7}}+1=\dfrac{\sqrt{3}+\sqrt{5}+\sqrt{7}}{\sqrt{7}}\)
\(A=\dfrac{\sqrt{\dfrac{5}{3}}}{\sqrt{\dfrac{7}{3}}+\sqrt{\dfrac{5}{3}}+1}+\dfrac{\sqrt{\dfrac{7}{5}}}{\sqrt{\dfrac{3}{5}}+\sqrt{\dfrac{7}{5}}+1}+\dfrac{\sqrt{\dfrac{3}{7}}}{\sqrt{\dfrac{5}{7}}+\sqrt{\dfrac{3}{7}}+1}\)
\(A=\dfrac{\sqrt{5}}{\sqrt{3}+\sqrt{5}+\sqrt{7}}+\dfrac{\sqrt{7}}{\sqrt{7}+\sqrt{5}+\sqrt{3}}+\dfrac{\sqrt{3}}{\sqrt{7}+\sqrt{5}+\sqrt{3}}=1\)
ST
0
LT
1
31 tháng 7 2020
Trả lời:
\(\frac{4}{\sqrt{7}-\sqrt{3}}+\frac{6}{3+\sqrt{3}}+\frac{\sqrt{7}-7}{\sqrt{7}-1}\)
\(=\frac{4.\left(\sqrt{7}+\sqrt{3}\right)}{7-3}+\frac{6.\left(3-\sqrt{3}\right)}{9-3}-\frac{7-\sqrt{7}}{\sqrt{7}-1}\)
\(=\frac{4.\left(\sqrt{7}+\sqrt{3}\right)}{4}+\frac{6.\left(3-\sqrt{3}\right)}{6}-\frac{\sqrt{7}.\left(\sqrt{7}-1\right)}{\sqrt{7}-1}\)
\(=\sqrt{7}+\sqrt{3}+3-\sqrt{3}-\sqrt{7}\)
\(=3\)
Học tốt
9 tháng 6 2020
Ta có: \(A=\frac{-\left(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\right)}{\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}}\)
\(=\frac{-\left(\sqrt{3-2\cdot\sqrt{3}\cdot2+4}+\sqrt{3+2\cdot\sqrt{3}\cdot2+4}\right)}{\sqrt{3-2\cdot\sqrt{3}\cdot2+4}-\sqrt{3+2\cdot\sqrt{3}\cdot2+4}}\)
\(=\frac{-\left(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{\left(\sqrt{3}+2\right)^2}\right)}{\sqrt{\left(\sqrt{3}-2\right)^2}-\sqrt{\left(\sqrt{3}+2\right)^2}}\)
\(=\frac{-\left(\left|\sqrt{3}-2\right|+\left|\sqrt{3}+2\right|\right)}{\left|\sqrt{3}-2\right|-\left|\sqrt{3}+2\right|}\)
\(=\frac{-\left(2-\sqrt{3}+\sqrt{3}+2\right)}{2-\sqrt{3}-\sqrt{3}-2}\)
\(=\frac{-4}{-2\sqrt{3}}=\frac{2\sqrt{3}}{3}\)
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