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b)=3^1+(3^2+3^3+3^4)+(3^5+3^6+3^7)+....+(3^58+3^59+3^60)
=3^1+(3^2.1+3^2.3+3^2.9)+(3^5.1+3^5.3+3^5.9)+......+(3^58.1+3^58.3+3^58.9)
=3^1+3^2.(1+3+9)+3^5.(1+3+9)+.....+3^58.(1+3+9)
=3+3^2.13+3^5.13+.........+3^58.13
=3.13.(3^2+3^5+....+3^58)
vi tich tren co thua so 13 nen tich do chia het cho 13
=
bai1
a) A=(31+32)+(33+34)+...+(359+360)
=(3^1.1+3^1.3)+...+(3^59.1+3^59.2)
=3^1.(1+3)+...+3^59.(1+3)
=3^1.4+....+3^59.4
=4.(3^1+...+3^59)
vi tich tren co thua so 4 nen tich do chia het cho 4
Phân tích ra \(6^9\cdot2^{10}+12^{10}=\left(2\cdot3\right)^9\cdot2^{10}+\left(3\cdot4\right)^{10}=2^9\cdot3^9\cdot2^{10}+3^{10}\cdot\left(2^2\right)^{10}=2^{19}\cdot3^9+\left(-3\right)^{10}\cdot2^{20}\)
\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}>\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}=\frac{15}{15}=1\)
\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< \frac{3}{9}+\frac{3}{9}+\frac{3}{9}+\frac{3}{9}+\frac{3}{9}=\frac{15}{9}< \frac{18}{9}=2\)
Suy ra đpcm.
Ta có: S =3/10+3/11+3/12+3/13+3/14 = 3.(1/10+1/11+1/12+1/13+1/14) > 3.(1/15 + 1/15 + 1/15 + 1/15 + 1/15) = 3.5/15 = 1 => S > 1 (1)
S=3/10+3/11+3/12+3/13+3/14 = 3.(1/10+1/11+1/12+1/13+1/14) < 3.(1/10 + 1/10 + 1/10 + 1/10 + 1/10) = 3.5/10 = 3/2<2 =>S <2 (2)
Từ (1) va (2)
=> 1 < S < 2 (đpcm).
Chúc bạn học tập tốt :)
- 3/10 +3/11 +3/12 +3/13 +3/14 > 3/15 + 3/15 +3/15 +3/15 +3/15 = 15/15 =1
S < 3/10 +3/10 +3/10 + 3/10 + 3/10 = 15/ 10 < 20/10 =2
Bài 1:
a: \(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4\cdot55⋮11\)
b: \(10^9+10^8+10^7\)
\(=10^7\left(10^2+10+1\right)=10^7\cdot111⋮111\)
S = 3/10 + 3/11 + 3/12 + 3/13 + 3/14 < 3/10 + 3/10 + 3/10 + 3/10 + 3/10
=> S < 5 x 3/10
=> S < 1,5
=> S < 2
S = 3/10 + 3/11 + 3/12 + 3/13 + 3/14 > 3/14 + 3/14 + 3/14 + 3/14 + 3/14
=> S > 5 x 3/14
=> S > 1,07.......
Mà 1 < 1,07 < S < 1,5 < 2
=> 1 < S < 2
\(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}>\frac{3}{14}+\frac{3}{14}+\frac{3}{14}+\frac{3}{14}+\frac{3}{14}\)
\(S>\frac{3}{14}.5\)
\(S>\frac{15}{14}>1\)(1)
\(S=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}\)
\(S< 5.\frac{3}{10}\)
\(S< \frac{15}{10}< 2\)(2)
Từ (1) và (2) => 1 < S < 2 => S không là số nguyên tố (đpcm)
Ta có:
\(10^{12}=1...0\)
\(\Rightarrow10^{12}+2=1..0+2=1...2\)
Mà:
\(1...2=1+0+...+0+2=3\) ⋮ 3
\(\Rightarrow10^{12}+2\) ⋮ 3
Ta có :
\(10^{12}+2=100...0\left(12.số0\right)+2\) có tổng các chữ số là \(1+0+...+0+2=3⋮3\) ( 12 chữ số 0)
\(\Rightarrow10^{12}+2⋮3\)