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1) Ta có: \(3\sqrt{12}+\dfrac{1}{2}\sqrt{48}-\sqrt{27}\)
\(=3\cdot2\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}-3\sqrt{3}\)
\(=6\sqrt{3}+2\sqrt{3}-3\sqrt{3}\)
\(=5\sqrt{3}\)
2) Ta có: \(\dfrac{2}{\sqrt{3}-5}\)
\(=\dfrac{2\left(\sqrt{3}+5\right)}{\left(\sqrt{3}-5\right)\left(\sqrt{3}+5\right)}\)
\(=\dfrac{2\left(\sqrt{3}+5\right)}{3-25}\)
\(=\dfrac{-2\left(\sqrt{3}+5\right)}{22}\)
\(=\dfrac{-\sqrt{3}-5}{11}\)
3) Ta có: \(\sqrt{\dfrac{2}{5}}\)
\(=\dfrac{\sqrt{2}}{\sqrt{5}}\)
\(=\dfrac{\sqrt{2}\cdot\sqrt{5}}{5}\)
\(=\dfrac{\sqrt{10}}{5}\)
Nếu em thấy các câu hỏi do lag mà bị gửi đúp (tức là rất nhiều câu hỏi giống nhau xuất hiện cùng 1 chỗ) thì xóa giúp mình nhé cho đỡ vướng. Nhưng nhớ để lại 1 câu. Cảm ơn em.
Lời giải:
\(\sqrt{\frac{(1+\sqrt{2})^3}{27}}=\sqrt{\frac{(1+\sqrt{2})^3}{3^3}}=\sqrt{\frac{3(1+\sqrt{2})^3}{3^4}}\)
\(=\frac{(1+\sqrt{2})\sqrt{3+3\sqrt{2}}}{9}\)
\(ab\sqrt{\frac{1}{a}+\frac{1}{b}}=\sqrt{(ab)^2(\frac{1}{a}+\frac{1}{b})}=\sqrt{ab^2+a^2b}\)
\(\sqrt{\dfrac{1}{600}}\)=\(\sqrt{\dfrac{1}{10^2\cdot6}}\)=\(\sqrt{\dfrac{1\cdot6}{10^2\cdot6\cdot6}}\)=\(\dfrac{\sqrt{6}}{60}\)
\(\sqrt{\dfrac{11}{540}}\)=\(\sqrt{\dfrac{11\cdot540}{540\cdot540}}\)=\(\dfrac{\sqrt{5940}}{540}\)=\(\dfrac{\sqrt{165}}{90}\)
\(\sqrt{\dfrac{3}{50}}\)=\(\sqrt{\dfrac{3\cdot50}{50\cdot50}}\)=\(\dfrac{\sqrt{150}}{50}\)=\(\dfrac{\sqrt{6}}{10}\)
\(\sqrt{\dfrac{5}{98}}\)=\(\sqrt{\dfrac{5\cdot98}{98\cdot98}}=\dfrac{\sqrt{490}}{98}=\dfrac{\sqrt{10}}{14}\)
\(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\dfrac{3-\sqrt{3}}{9}\)
\(\sqrt{\dfrac{1}{600}}=\dfrac{\sqrt{6}}{60}\)
\(\sqrt{\dfrac{11}{540}}=\dfrac{\sqrt{165}}{90}\)
\(\sqrt{\dfrac{3}{50}}=\dfrac{\sqrt{6}}{10}\)
\(\sqrt{\dfrac{5}{98}}=\dfrac{\sqrt{10}}{14}\)
\(\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{27}}=\dfrac{3-\sqrt{3}}{9}\)
EM thử thôi, ko chắc đâu ạ:( Sai thì xin thông cảm cho ạ.
1) \(\sqrt{\frac{2}{3-\sqrt{5}}}=\sqrt{\frac{2\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}}=\sqrt{\frac{6+2\sqrt{5}}{4}}=\frac{\sqrt{6+2\sqrt{5}}}{2}\)
2) \(\sqrt{\frac{a-4}{2\left(\sqrt{a}-2\right)}}=\sqrt{\frac{\left(a-4\right)\left(\sqrt{a}+2\right)}{2\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}}\)
\(=\sqrt{\frac{\left(a-4\right)\left(\sqrt{a}+2\right)}{2\left(a-4\right)}}\)
3) \(\sqrt{\frac{1}{a\left(1-\sqrt{3}\right)}}=\sqrt{\frac{1+\sqrt{3}}{a\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}}=\sqrt{\frac{1+\sqrt{3}}{a\left(1-3\right)}}=\sqrt{-\frac{1+\sqrt{3}}{2a}}\)
4) \(\sqrt{\frac{a}{4-2\sqrt{3}}}=\sqrt{\frac{a\left(4+2\sqrt{3}\right)}{\left(4-2\sqrt{3}\right)\left(4+2\sqrt{3}\right)}}=\sqrt{\frac{4a+2a\sqrt{3}}{16-12}}=\sqrt{\frac{4a+2a\sqrt{3}}{4}}=\frac{\sqrt{4a+2a\sqrt{3}}}{2}\)
a, \(\sqrt{\frac{1}{60}}=\frac{\sqrt{1}}{\sqrt{60}}=\frac{\sqrt{1}.\sqrt{60}}{\sqrt{60}.\sqrt{60}}=\frac{\sqrt{60}}{60}=\frac{2.\sqrt{15}}{2.30}=\frac{\sqrt{15}}{30}\)
c, \(\frac{1}{2-\sqrt{3}}=\frac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}\)
d, \(\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}=\frac{\left(\sqrt{7}-\sqrt{3}\right)^2}{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}=\frac{7-2\sqrt{21}+3}{7-3}=\frac{10-2\sqrt{21}}{4}\)
Khử mẫu biểu thức chứa căn ms đúng
\(\sqrt{\frac{\left(1+\sqrt{2}\right)^3}{27}}=\sqrt{\frac{\left(1+\sqrt{2}\right)^2\cdot\left(1+\sqrt{2}\right)}{3^2\cdot3}}=\frac{1+\sqrt{2}}{3}\cdot\sqrt{\frac{1+\sqrt{2}}{3}}\)
\(=\frac{1+\sqrt{2}}{3}\cdot\frac{\sqrt{3\cdot\left(1+\sqrt{2}\right)}}{3}=\frac{1+\sqrt{2}}{9}\cdot\sqrt{3+3\sqrt{2}}\)
\(\sqrt{\frac{\left(1-\sqrt{3}\right)^2}{27}}=\sqrt{\frac{4-2\sqrt{3}}{27}}=0,1408832436\)
ta có :
\(-4\sqrt{\frac{\sqrt{3}-1}{2+\sqrt{3}}}=-4\sqrt{\frac{\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)}{\left(2+\sqrt{3}\right)^2}}=-\frac{4\sqrt{1+\sqrt{3}}}{\left(2+\sqrt{3}\right)}=-\frac{2\sqrt{4+4\sqrt{3}}}{\left(2+\sqrt{3}\right)}\)
\(=-\frac{2\sqrt{\left(1+\sqrt{3}\right)^2}}{\left(2+\sqrt{3}\right)}=-\frac{2\left(1+\sqrt{3}\right)}{\left(2+\sqrt{3}\right)}=-\frac{2\left(1+\sqrt{3}\right)\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(\left(2-\sqrt{3}\right)\right)}=-2\left(\sqrt{3}-1\right)\)
\(=2-2\sqrt{3}\)
\(-4\sqrt{\frac{\sqrt{3}-1}{2+\sqrt{3}}}\)
\(-4\sqrt{\frac{\left(\sqrt{3}-1\right)\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}\)
\(=-4\sqrt{\frac{2\sqrt{3}-2-3+\sqrt{3}}{4-3}}\)
\(=-\left(4\sqrt{3\sqrt{3}-5}\right)\)
\(=-\sqrt{48\sqrt{3}-80}\)