27.(62+37)+73.(43+56)

= 27.99 + 73 . 99

= 99 .( 27+73)

= 99. 100

= 9900

`27.(62 + 37) + 73.(43+56)`

`=27.99 + 73 . 99`

`=99 . (27 + 73)`

`=99 . 100`

`=9900`

`#LeMichael`

\(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\dfrac{11.3^{22+7}-\left(3^2\right)^{15}}{2^2.3^{14.2}}=\dfrac{11.3^{29}-3^{30}}{4.3^{28}}\)

\(=\dfrac{3^{28}\left(11.3-3^2\right)}{4.3^{28}}=\dfrac{33-9}{4}=\dfrac{24}{4}=6\)

Đề là gì bạn nhỉ?

cho mình đề bài ạ

Bạn lm sai ở mấy bước cuối nhé!

Bài làm:

`7.2^3-72:3^2`

`=7.8-72:9`

`=7.8-8`

`=8.(7-1)`

`=8.6=48`

hình như là sai 

\(45:\dfrac{7}{18}+50%-1,25\)

\(=45. \dfrac{18}{7}+\dfrac{1}{2}-\dfrac{5}{4}\)

\(=\dfrac{810}{7}+\dfrac{1}{2}-\dfrac{5}{4}\)

\(=\dfrac{1627}{14}-\dfrac{5}{4}\)

\(=\dfrac{3219}{28}\)

   45: \(\dfrac{7}{18}\) + 50% - 1,25

= 45 x \(\dfrac{18}{7}\) + \(\dfrac{50}{100}\) - \(\dfrac{125}{100}\)

   =   \(\dfrac{810}{7}\)- \(\dfrac{75}{100}\)

= \(\dfrac{81000}{700}\) - \(\dfrac{525}{700}\)

=\(\dfrac{80475}{700}\)

=\(\dfrac{3219}{28}\)

`b)x.2010-x=2010.2011-2011`

`=>x(2010-1)=2011(2010-1)`

`=>x=2011`

__________________________

`c)x:1+x.2,7+x.6,3=70`

`=>x.1+x.2,7+x.6,3=70`

`=>x(1+2,7+6,3)=70`

`=>x.10=70`

`=>x=70:10=7`

4 . 2^x - 3 = 125

4 . 2^x = 125 + 3 

4 . 2^x = 128

2^x = 128 : 4

2^x = 32

2^x = 2⁵

=> x = 5.

\(4\cdot2^x-3=125\)

\(\Leftrightarrow4\cdot2^x=128\)

\(\Leftrightarrow2^x=32\)

\(\Leftrightarrow2^x=2^5\)

\(\Rightarrow x=5\)

a, A = 2003.2003 = 2003 , 2002 + 2003 > 2002.2003 + 2002 

2002.2003 + 2002 = 2002.2004

Vậy 2003.2003 > 2002.2004

b, A = 1990.2010 và B = 2170.2000

A =   (2000- 10).(2000 + 10) = 2000.2000 -100<2000.2170

vậy 1990.2010 < 2000.2170

`3x-15=30`

`3x=30+15`

`3x=45`

`x=45:3`

`x=15`

______________________________

`165-(x+15)=82`

`x+15=165-82`

`x+15=83`

`x=83-15`

`x=68`

__________________________________

`(x+20)-17=43`

`x+20=43+17`

`x+20=60`

`x=60-20`

`x=40`

_______________________________

`35 +x - 25 = 602`

`35+x=602 + 25`

`35+x = 627`

`x=627 - 35`

`x=592`

________________________________

`(x-3).(2x+4).(x-6)=0`

`=>` \(\left[{}\begin{matrix}x-3=0\\2x+4=0\\x-6=0\end{matrix}\right.\)

`=>` \(\left[{}\begin{matrix}x=3\\x=-2\\x=6\end{matrix}\right.\)

____________________________________

`#LeMichael`

3x-15=30

3x=30+15

3x=45

x=45:3

x=15