đổi 32,4km/giờ=9m/giây

54 km / giờ =15m/giây

chiều dài toa tàu là:

(15-9)x25=150(m)

         Đ/s:...

490-(150:3*4-7)=490-(50*4-7)=490-(200-7)=290+7=297

=490-{ [ 150 / 3 * 4] - 7 } = 490 - { [ 50 * 4 ] - 7 } = 490 - 200 + 7 = 297

= 2⁵ . (2³)⁴ = 2⁵⁺¹²= 2¹⁷

= (5²)⁶.(5³)³ = 5¹².5⁹=5¹²⁺⁹= 5²¹

= 144

=1

= 0

 = 400

\(12^2=144\)

\(1^{100}=1\)

\(100^1=100\)

\(0^{2018}=0\)

\(20^2=400\)

a)\(315-64:2^3\)

\(=315-2^6:2^3\)

\(=315-8=307\)

b)\(7^2.33+7^2.67\)

\(=7^2\left(33+67\right)\)

\(=7^2.100=4900\)

a) \(315-64\div2^3=315-2^6\div2^3=315-2^{6-3}=315-2^3=315-8=307\)

b) \(7^2.33+7^2.67=7^2.\left(33+67\right)=7^2.100=49.100=4900\)

\(A=\frac{2017.2018-1}{2017.2018}=1-\frac{1}{2017.2018}\)

\(B=\frac{2018.2019-1}{2018.2019}=1-\frac{1}{2018.2019}\)

Có \(\frac{1}{2017.2018}>\frac{1}{2018.2019}\)

\(\Rightarrow A< B\)

\(A=\frac{2017.2018-1}{2017.2018}=1-\frac{1}{2017.2018}\)

\(B=\frac{2018.2019-1}{2018.2019}=1-\frac{1}{2018.2019}\)

Do  \(\frac{1}{2017.2018}>\frac{1}{2018.2019}\)nên  \(1-\frac{1}{2017.2018}< 1-\frac{1}{2018.2019}\)

Vậy  \(A< B\)

15.360+150.2.27+1500

=15.36.10+150.54+150.10

=150.36+150.54+150.10

=150.(36+54+10)

=150.100

=15000

cứ chép thế ko cần hiểu

15 . 360 + 150 . 2 . 27 + 1500

= 15 . ( 10 . 36 ) + 150 . ( 2 . 27 ) + 150 . 10

= 15 . 10 . 36 + 150 . 54 + 150 . 10

= 150 . 36 + 150 . 54 + 150 . 10

= 150 . ( 36 + 54 + 10 )

= 150 . 100

= 15 000  

 Ta có : 12 = 2² . 3             36 = 2² . 3²             48 = 2⁴ . 3 

=> BCNN(12;36;48) = 2⁴ . 3² = 144

=> B(144) = { 0 ;144 ;288 ;432 ;576 ;720 ;...}

=> BC(12;36;48) = { 0 ; 144 ; 288 ; 432 ; 576 ; 720 ; ...}