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1) \(\sqrt{1\dfrac{9}{16}}=\sqrt{\dfrac{25}{16}}=\dfrac{5}{4}\)
2) \(\dfrac{\sqrt{12.5}}{0.5}=\sqrt{\dfrac{12.5}{0.25}}=5\sqrt{2}\)
3) \(\sqrt{\dfrac{25}{64}}=\dfrac{5}{8}\)
4) \(\dfrac{\sqrt{230}}{\sqrt{2.3}}=\sqrt{\dfrac{230}{2.3}}=\sqrt{100}=10\)
5) \(\left(\sqrt{\dfrac{2}{3}}+\sqrt{\dfrac{50}{3}}-\sqrt{24}\right)\cdot\sqrt{6}\)
\(=\left(\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{5\sqrt{2}}{\sqrt{3}}-2\sqrt{6}\right)\cdot\sqrt{6}\)
\(=\left(\dfrac{6\sqrt{2}}{\sqrt{3}}-2\sqrt{6}\right)\cdot\sqrt{6}\)
\(=0\cdot\sqrt{6}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\sqrt{\dfrac{2}{2-\sqrt{3}}}=\sqrt{\dfrac{2\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)\(=\sqrt{2\left(2+\sqrt{3}\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
b)\(\sqrt{\dfrac{2}{3}}-\sqrt{24}+2\sqrt{\dfrac{3}{8}}+\dfrac{1}{6}=\dfrac{\sqrt{6}}{3}-\sqrt{2^2.6}+\dfrac{2\sqrt{24}}{8}+\dfrac{1}{6}\)
\(=\dfrac{\sqrt{6}}{3}-2\sqrt{6}+\dfrac{\sqrt{2^2.6}}{4}+\dfrac{1}{6}=\dfrac{-5\sqrt{6}}{3}+\dfrac{2\sqrt{6}}{4}+\dfrac{1}{6}\)
\(=\dfrac{-7\sqrt{6}}{6}+\dfrac{1}{6}\)
Nhầm xíu
\(\sqrt{\dfrac{2}{2-\sqrt{3}}}-\sqrt{\dfrac{2}{2+\sqrt{3}}}=\sqrt{\dfrac{2\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}-\sqrt{\dfrac{2\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{2\left(2+\sqrt{3}\right)}-\sqrt{2\left(2-\sqrt{3}\right)}=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}+1-\left|\sqrt{3}-1\right|=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{\sqrt{12}-6}{\sqrt{8}-\sqrt{24}}-\dfrac{3+\sqrt{3}}{\sqrt{3}}+\dfrac{4}{1-\sqrt{7}}\)
\(=\dfrac{2\sqrt{3}\cdot\left(1-\sqrt{3}\right)}{2\sqrt{2}\cdot\left(1-\sqrt{3}\right)}-\dfrac{\sqrt{3}\cdot\left(\sqrt{3}+1\right)}{\sqrt{3}}+\dfrac{4\left(1+\sqrt{7}\right)}{\left(1-\sqrt{7}\right)\left(1+\sqrt{7}\right)}\)
\(=\dfrac{2\sqrt{3}}{2\sqrt{2}}-\left(\sqrt{3}+1\right)-\dfrac{4\left(1+\sqrt{7}\right)}{1-7}\)
\(=\dfrac{\sqrt{3}}{\sqrt{2}}-\sqrt{3}-1-\dfrac{4\left(1+\sqrt{7}\right)}{-6}\)
\(=\dfrac{2\sqrt{3}}{2}-\sqrt{3}-1+\dfrac{2+2\sqrt{7}}{3}\)
\(=\dfrac{6\sqrt{3}-6\left(\sqrt{3}+1\right)+2\left(2+2\sqrt{7}\right)}{6}\)
\(=\dfrac{6\sqrt{3}-6\sqrt{3}-6+4+4\sqrt{7}}{6}\)
\(=\dfrac{4\sqrt{7}-2}{6}\)
\(=\dfrac{2\sqrt{7}-1}{3}\)
\(=\dfrac{\sqrt{12}\left(1-\sqrt{3}\right)}{2\sqrt{2}\left(1-\sqrt{3}\right)}-\sqrt{3}-1-\dfrac{4\left(\sqrt{7}+1\right)}{6}\)
\(=\dfrac{\sqrt{6}}{2}-\sqrt{3}-1-\dfrac{2}{3}\left(\sqrt{7}+1\right)\)
\(=\dfrac{\sqrt{6}}{2}-\sqrt{3}-1-\dfrac{2}{3}\sqrt{7}-\dfrac{2}{3}\)
\(=\dfrac{1}{2}\sqrt{6}-\sqrt{3}-\dfrac{2}{3}\sqrt{7}-\dfrac{5}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)
\(=\sqrt{5}+\dfrac{\sqrt{5}}{2}\)
\(=\dfrac{2\sqrt{5}}{2}+\dfrac{\sqrt{5}}{2}\)
\(=\dfrac{3\sqrt{5}}{2}\)
\(\left(\sqrt{3}+1\right)\cdot\dfrac{\sqrt{3}-3}{2\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\cdot\dfrac{\sqrt{3}\left(1-\sqrt{3}\right)}{2\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\cdot\dfrac{1-\sqrt{3}}{2}\)
\(=\dfrac{\left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)}{2}\)
\(=\dfrac{1-3}{2}\)
\(=-1\)
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\(1,=2\sqrt{3}-3\sqrt{3}+4\sqrt{3}=3\sqrt{3}\\ 2,=\left(2\sqrt{6}+2\sqrt{5}-4\sqrt{5}\right):5=\dfrac{2\sqrt{6}}{5}-\dfrac{2\sqrt{5}}{5}\\ 3,=6\sqrt{3}-\dfrac{4\sqrt{3}}{3}-4\sqrt{3}-\dfrac{5\sqrt{3}}{3}=2\sqrt{3}-\dfrac{9\sqrt{3}}{3}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\\ 4,Sửa:\dfrac{1}{\sqrt{5}-\sqrt{3}}-\dfrac{1}{\sqrt{5}+\sqrt{3}}\\ =\dfrac{\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\)
1) \(=2\sqrt{3}-3\sqrt{3}+4\sqrt{3}=3\sqrt{3}\)
2) \(=\left(2\sqrt{6}+2\sqrt{5}-4\sqrt{5}\right)=\dfrac{2\sqrt{6}}{5}+\dfrac{2\sqrt{5}}{5}-\dfrac{4\sqrt{5}}{5}\)
3) \(=6\sqrt{3}-\dfrac{4\sqrt{3}}{3}-4\sqrt{3}-\dfrac{5\sqrt{3}}{3}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\)
4) \(=\dfrac{\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}}{5-3}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b)
\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)
\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)=6-121=-115\)
a)
\(\dfrac{\left(3\sqrt{3}+5\sqrt{2}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)
\(=\dfrac{\left(3\sqrt{3}+5\sqrt{2}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\)
\(=\dfrac{\left(3\sqrt{3}+5\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{5\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\dfrac{9-3\sqrt{6}+5\sqrt{6}-10}{5}=\dfrac{-1+2\sqrt{6}}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`c)1/(2sqrt2)-3/2sqrt{4,5}+2/5sqrt{50}`
`=1/(2sqrt2)-3/2sqrt{9/2}+2/5sqrt{25.2}`
`=1/(2sqrt2)-9/(2sqrt2)+2sqrt2`
`=2sqrt2-8/(2sqrt2)`
`=2sqrt2-sqrt2=sqrt2`
`d)4/(3+sqrt5)-8/(1+sqrt5)+15/sqrt5`
`=(4(3-sqrt5))/(9-5)-(8(sqrt5-1))/(5-1)+3sqrt5`
`=3-sqrt5-2(sqrt5-1)+3sqrt5`
`=3+3sqrt5-3sqrt5+2=5`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=5\sqrt{2}-\dfrac{3}{2}\cdot4\sqrt{2}-\dfrac{1}{3}\cdot6\sqrt{2}+8=-3\sqrt{2}+8\)
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A: \(A=\sqrt{9}-3\sqrt{\dfrac{50}{9}}+3\sqrt{8}-\sqrt[3]{27}\)
\(=3-3\cdot\dfrac{5\sqrt{2}}{3}+6\sqrt{2}-3\)
\(=-5\sqrt{2}+6\sqrt{2}=\sqrt{2}\)
b: \(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}-1}-6\cdot\sqrt{\dfrac{16}{3}}\)
\(=\left|2-\sqrt{3}\right|+\dfrac{2\left(\sqrt{3}+1\right)}{3-1}-6\cdot\dfrac{4}{\sqrt{3}}\)
\(=2-\sqrt{3}+\sqrt{3}+1-4\sqrt{3}\)
\(=3-4\sqrt{3}\)
\(A=\sqrt{9}-3\sqrt{\dfrac{50}{9}}+3\sqrt{8}-\sqrt[3]{27}\\ =3-3\cdot\dfrac{1}{3}\sqrt{25\cdot2}+3\sqrt{4\cdot2}-3\\ =3-1\cdot5\sqrt{2}+3\cdot2\sqrt{2}-3\\ =3-5\sqrt{2}+6\sqrt{2}-3\\ =\sqrt{2}\)
\(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}-1}-6\sqrt{\dfrac{16}{3}}\\ =\left|2-\sqrt{3}\right|+\dfrac{2\left(\sqrt{3}+1\right)}{3-1}-6\cdot\dfrac{4\sqrt{3}}{3}\\ =2-\sqrt{3}+\sqrt{3}+1-8\sqrt{3}\\ =3-8\sqrt{3}\)
\(=\dfrac{2\sqrt{6}}{3}-2\sqrt{6}-\dfrac{5}{3}\sqrt{6}=-3\sqrt{6}\)