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`#3107.101107`
a)
`2/5 \sqrt{25} - 1/2 \sqrt{4}`
`= 2/5 * \sqrt{5^2} - 1/2 * \sqrt{2^2}`
`= 2/5*5 - 1/2*2`
`= 2 - 1`
`= 1`
b)
`0,5*\sqrt{0,09} + 5*\sqrt{0,81}`
`= 0,5*\sqrt{(0,3)^2} + 5*\sqrt{(0,9)^2}`
`= 0,5*0,3 + 5*0,9`
`= 0,15 + 4,5`
`= 4,65`
c)
`2/5\sqrt{25/36} - 5/2\sqrt{4/25}`
`= 2/5*\sqrt{(5^2)/(6^2)} - 5/2*\sqrt{(2^2)/(5^2)}`
`= 2/5*5/6 - 5/2*2/5`
`= 1/3 - 1`
`= -2/3`
d)
`-2 \sqrt{(-36)/(-16)} + 5 \sqrt{(-81)/(-25)}`
`= -2*\sqrt{36/16} + 5*\sqrt{81/25}`
`= -2*\sqrt{(6^2)/(4^2)} + 5*\sqrt{(9^2)/(5^2)}`
`= -2*6/4 + 5*9/5`
`= -3 + 9`
`= 6`
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(D=\left(\sqrt{57}+3\sqrt{6}+\sqrt{38}+6\right)\left(\sqrt{57}-3\sqrt{6}-\sqrt{38}+6\right)\)
\(=\left(\sqrt{57}+6\right)^2-\left(3\sqrt{6}+\sqrt{38}\right)^2\)
\(=57+12\sqrt{57}+36-\left(54+12\sqrt{57}+38\right)\)
\(=93-92=1\)
Vậy : \(D=1\)
\(D=\left(\sqrt{57}+3\sqrt{6}+\sqrt{38}+6\right)\left(\sqrt{57}-2\sqrt{6}-\sqrt{38}+6\right)\)
\(=\left(\sqrt{57}+6\right)^2-\left(3\sqrt{6}+\sqrt{38}\right)^2\)
\(=\left(93+12\sqrt{57}\right)-\left(92+12\sqrt{57}\right)\)
\(=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
26, đặt bthuc là A suy ra A2=4+4+2\(\sqrt{16-\left(10+2\sqrt{5}\right)}\) suy ra A2=8+2(\(\sqrt{5}\) -1) suy ra A=\(\sqrt{6+2\sqrt{5}}\)=\(\sqrt{5}\)+1
40, tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt{16}\cdot\sqrt{25}+\sqrt{196}:\sqrt{49}\)
\(=\sqrt{16\cdot25}+\sqrt{196:49}\)
\(=20+2=22\)
b) \(36:\sqrt{2\cdot3^2\cdot18}-\sqrt{169}\)
\(=36:\sqrt{324}-\sqrt{169}\)
\(=36:18-13=2-13=-11\)
c) \(\sqrt{\sqrt{81}}\)
\(=\sqrt{9}=3\)
d) \(\sqrt{3^2+4^2}\)
\(=\sqrt{9+16}=\sqrt{25}=5\)
a) \(\sqrt{16}.\sqrt{25}+\sqrt{196}\div\sqrt{49}\)
\(=4.5+14:7\)
\(=20+2=22\)
b) \(36:\sqrt{2.3^2.18}-\sqrt{169}\)
\(=36:18-13=-11\)
c) \(\sqrt{\sqrt{81}}=\sqrt{9}=3\)
d) \(\sqrt{3^2+4^2}=\sqrt{25}=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt{36-6\times2\sqrt{5}+5-5}\)
=\(\sqrt{\left(6-\sqrt{5}\right)^2-5}\)=\(\sqrt{\left(6-\sqrt{5}-\sqrt{5}\right)\left(6-\sqrt{5}+\sqrt{5}\right)}\)
=\(\sqrt{\left(5-2\sqrt{5}+1\right)\times6}\)
=\(\sqrt{\left(\sqrt{5}-1\right)^2\times6}\)
=(\(\sqrt{5}-1\))\(\times\)\(\sqrt{6}\)
Câu b muộn rùi nghỉ đây bạn tự nghĩ đi dễ mà