1: \(\sqrt{147}+\sqrt{54}-4\sqrt{27}\)

\(=7\cdot\sqrt3+3\sqrt6-4\cdot3\sqrt3\)

\(=3\sqrt6+7\sqrt3-12\sqrt3=3\sqrt6-5\sqrt3\)

2: \(\sqrt{28}-4\cdot\sqrt{63}+7\cdot\sqrt{112}\)

\(=2\sqrt7-4\cdot3\sqrt7+7\cdot4\sqrt7\)

\(=2\sqrt7-12\sqrt7+28\sqrt7=18\sqrt7\)

3: \(\sqrt{49}-5\cdot\sqrt{28}+\frac12\cdot\sqrt{63}\)

\(=7-5\cdot2\sqrt7+\frac12\cdot3\sqrt7=7-10\sqrt7+1,5\sqrt7=7-8,5\cdot\sqrt7\)

4: \(\left(2\sqrt6-4\sqrt3-\frac14\sqrt8\right)\cdot3\sqrt6\)

\(=6\cdot\sqrt{36}-12\sqrt{18}-\frac14\cdot2\cdot\sqrt2\cdot3\sqrt6\)

\(=36-36\sqrt2-\frac32\sqrt{12}=36-36\sqrt2-3\sqrt3\)

6: \(\left(\sqrt{48}-3\sqrt{27}-\sqrt{147}\right):\sqrt3=\sqrt{\frac{48}{3}}-3\cdot\sqrt{\frac{27}{3}}-\sqrt{\frac{147}{3}}\)

\(=\sqrt{16}-3\cdot\sqrt9-\sqrt{49}=4-3\cdot3-7\)

=-3-9

=-12

5: \(\left(2\cdot\sqrt{1\frac{9}{16}}-5\cdot\sqrt{5\frac{1}{16}}\right):\sqrt{16}\)

\(=\left(2\cdot\sqrt{\frac{25}{16}}-5\cdot\sqrt{\frac{81}{16}}\right):\sqrt{16}\)

\(=\left(2\cdot\frac54-5\cdot\frac94\right):4=\frac{10-45}{4\cdot4}=\frac{-35}{16}\)

7: \(\left(\sqrt{50}-3\sqrt{49}\right):\sqrt2-\sqrt{162}:\sqrt2\)

\(=\left(5\sqrt2-21\right):\sqrt2-9\sqrt2:\sqrt2\)

\(=5-\frac{21}{\sqrt2}-9=-4-\frac{21\sqrt2}{2}=\frac{-8-21\sqrt2}{2}\)

8: \(\left(2\cdot\sqrt{1\frac{9}{10}}-\sqrt{5\frac{1}{10}}\right):\sqrt{10}=\left(2\cdot\sqrt{\frac{19}{10}}-\sqrt{\frac{51}{10}}:\sqrt{10}\right)\)

\(=2\cdot\frac{\sqrt{19}}{10}-\frac{\sqrt{51}}{10}=\frac{2\sqrt{19}-\sqrt{51}}{10}\)

9: \(2\cdot\sqrt{\frac{16}{3}}-3\cdot\sqrt{\frac{1}{27}}-6\cdot\sqrt{\frac{4}{75}}\)

\(=2\cdot\frac{4}{\sqrt3}-3\cdot\frac{1}{3\sqrt3}-6\cdot\frac{2}{5\sqrt3}\)

\(=\frac{8}{\sqrt3}-\frac{1}{\sqrt3}-\frac{12}{5\sqrt3}=\frac{7}{\sqrt3}-\frac{12}{5\sqrt3}=\frac{7\sqrt3}{3}-\frac{12\sqrt3}{15}=\frac{35\sqrt3-12\sqrt3}{15}=\frac{23\sqrt3}{15}\)

10: \(2\sqrt{27}-6\cdot\sqrt{\frac43}+\frac35\cdot\sqrt{75}\)

\(=2\cdot3\sqrt3-6\cdot\frac{2}{\sqrt3}+\frac35\cdot5\sqrt3\)

\(=6\sqrt3-4\sqrt3+3\sqrt3=5\sqrt3\)

11: \(\frac{\sqrt{18}}{\sqrt2}-\frac{\sqrt{12}}{\sqrt3}=\sqrt9-\sqrt4=3-2=1\)

2:

a: A=1+2+2^2+2^3+2^4

=>2A=2+2^2+2^3+2^4+2^5

=>A=2^5-1

=>A=B

b: C=3+3^2+...+3^100

=>3C=3^2+3^3+...+3^101

=>2C=3^101-3

=>\(C=\dfrac{3^{101}-3}{2}\)

=>C=D

Ta có: 

\(\left\{\begin{matrix}5^{27}=\left(5^3\right)^9=125^9\\2^{63}=\left(2^7\right)^9=128^9\end{matrix}\right\}\Rightarrow5^{27}< 2^{63}\left(1\right)\)

\(\left\{\begin{matrix}2^{63}=\left(2^9\right)^7=512^7\\5^{28}=\left(5^4\right)^7=625^7\end{matrix}\right\}\Rightarrow2^{63}< 5^{28}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow5^{27}< 2^{63}< 5^{28}\) (đpcm)

a ) 37 x 27 + 63 x 27 

= ( 37 + 63 ) x 27

= 100 x 27

= 2700

b ) 1 + 2 + 3 + 4 + ... + 97 + 98 + 99

= ( 99 - 1 ) : 1 + 1 

= 99 x ( 99 + 1 ) : 2 

= 4950

c ) 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10

= ( 1 + 9 ) + ( 2 + 8 ) + ( 3 + 7 ) + ( 4 + 6 ) + ( 5 + 10 )

= 10 + 10 + 10 + 10 + 15

= 55

37 x 27 x 63 x27

=(37+63)x27

=100 x 27

nhầm là 105

bấm máy tính nhak bạn

cái này bấm máy tính là thầy mình k cho đâu =))) phải trình bày ra mà

c: Ta có: \(\dfrac{2}{5}\cdot\left[\left(\dfrac{3}{5}\right)^2:\left(-\dfrac{1}{5}\right)^2-7\right]\cdot\left(1000\right)^0\cdot\left|-\dfrac{11}{15}\right|\)

\(=\dfrac{2}{5}\cdot\left(\dfrac{9}{25}:\dfrac{1}{25}-7\right)\cdot1\cdot\dfrac{11}{15}\)

\(=\dfrac{2}{5}\cdot\dfrac{11}{15}\cdot2\)

\(=\dfrac{44}{75}\)

Cảm ơn bạn lần nữa! 

`4/7+4`

`=4/7+4/1`

`=4/7+28/7`

`=32/7`

__

`3+6/11`

`=33/11+6/11`

`=39/11`

__

`3-5/7`

`=3/1-5/7`

`=21/7-5/7`

`=16/7`

__

`21/9-2`

`=21/9-18/9`

`=3/9`

`=1/3`

__

`15/24+2`

`=15/24+48/24`

`=63/24`

`=21/16`

__

`63/45-20/25`

`=63/45-4/5`

`=63/45-36/45`

`=27/45`

`=9/15`

__

`3/4-2/8`

`=3/4-1/4`

`=2/4`

__

`6/7-5/8`

`=48/56-35/56`

`=13/56`

__

`37/45-5/9`

`=37/45-25/45`

`=12/45`

`=4/15`

__

`46/39-11/13`

`=46/39-33/39`

`=13/39`

`=1/2`

__

`5/12+3/4+1/3`

`=5/12+9/12+4/12`

`=14/12+4/12`

`=18/12`

`=3/2`

__

`1/2+3/7+11/14`

`=7/14+6/14+11/14`

`=13/14+11/14`

`=24/14`

`=12/7`

__

`7/10-(1/5+1/4)`

`=7/10-(4/20+5/20)`

`=7/10-9/20`

`=14/20-9/20`

`=5/20`

`=1/4`

__

`15/4-2/3-3/4`

`=(15/4-3/4)-2/3`

`=12/4-2/3`

`=3-2/3`

`=9/3-2/3`

`=7/3`