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A = \(\dfrac{21^2.14.125}{35^5.6}\)
= \(\dfrac{\left(3.7\right)^2.\left(2.7\right).5^3}{\left(5.7\right)^5.\left(2.3\right)}\)
= \(\dfrac{3^2.7^2.2.7.5^3}{5^5.7^5.2.3}\)
= \(\dfrac{3^2.7^3.2.5^3}{5^5.7^5.2.3}\)
= \(\dfrac{3}{5^2.7^2}\)
= \(\dfrac{3}{35^2}\)
@Pham Thi Thu Trang
B = \(\dfrac{2^{10}.13+2^{10}.65}{2^8.104}\)
= \(\dfrac{2^{10}\left(13+65\right)}{2^8.104}\)
= \(\dfrac{2.78}{104}\)
= \(\dfrac{3}{2}\)
@Pham Thi Thu Trang
a) Ta có 120a + 36b = 12.10a + 12.3b = 12(10a + 3b) \(⋮\)12
b) Ta có 57 - 56 + 55 = 55(52 - 5 + 1) = 55.21 \(⋮\)21
c) Ta có 52012 + 52013 + 52014 = 52012(1 + 5 + 52) = 52012.31 \(⋮31\)
d) Ta có 76 + 75 - 74 = 74(72 + 7 - 1) = 74.55 = 73.7.11.5 = 73.5.77 \(⋮\)77
a) Vì \(\hept{\begin{cases}120⋮12\\36⋮12\end{cases}\Rightarrow}\hept{\begin{cases}120a⋮12\\36b⋮12\end{cases}}\Rightarrow\left(120a+36b\right)⋮12\)
b) \(5^7-5^6+5^5=5^5\left(5^2-5+1\right)=5^5\left(25-6+1\right)=21.5^5⋮21\)
c)\(5^{2012}+5^{2013}+5^{2014}=5^{2012}\left(1+5+5^2\right)=5^{2012}\left(1+5+25\right)=31.5^{2012}⋮31\)
d)\(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4\left(49+7-1\right)=55.7^4=11.5.7^4⋮11\)
Dễ thấy : \(7^6+7^5-7^4⋮7\)
mà \(\left(11;7\right)=1\)
\(\Rightarrow7^6+7^5-7^4⋮77\)
\(5^{12}:5^8=5^4\)
\(7^9:7^6=7^3\)
\(5^5.5^3=5^8\)
\(4^5.2=2^{10}.2=2^{11}\)
Ta có : A = (7 + 72 + 73 ) + (74 + 75 + 76 ) + ( 77 + 78 + 79 ) + (710 +711 + 712 )
=7.(1 + 7 + 72 ) + 74 .(1 + 7 + 72 ) + 77 .(1 +7 +72 ) + 710 .(1 + 7 +72 )
= 57.( 7 + 74 + 77 + 710 ) chia hết cho 57.
Vậy số dư phép chia A cho 57 là 0
\(12^7.6^7=\left(12.6\right)^7=72^7=\text{10030613004288}\)
127 : 67
= (12 : 6)7
= 27
= 128