\(a=15\times23,45+860\times2,345-2345\\ a=15\times2345\times\dfrac{1}{100}+860\times2345\times\dfrac{1}{1000}-2345\\ a=2345\times\left(15\times\dfrac{1}{100}+860\times\dfrac{1}{1000}-1\right)\\ a=2345\times0,01=23,45\)

Gọi số có 3 chữ số cần tìm là \(\overline{abc}\)

Ta biết: 

\(\overline{3abc}=\overline{abc}\times25\\ 3000+\overline{abc}=\overline{abc}\times25\\ \overline{abc}\times25-\overline{abc}=3000\\ \overline{abc}\times24=3000\\ \overline{abc}=125\)

Số số hạng từ 2;4;6.......;100 là: (100-2):2+1= 50 (số)

tổng : (2+100)x50:2 +105 =2655

a) Gọi \(2^0+2^1+...+2^{2017}\) là A

\(A=2^0+2^1+2^2+...+2^{2017}\\ 2A=2+2^2+2^3+....+2^{2018}\\ 2A-A=\left(2+2^2+2^3+....+2^{2018}\right)-\left(2^0+2^1+2^2+...+2^{2017}\right)\\ A=2^{2018}-1=4^x-1\\ =>2^{2018}=4^x=>4^{1009}=4^x=>x=1009\)

\(b,3^x+3^{x+1}+3^{x+2}+3^{x+3}=360\\ 3^x+3^x\cdot3+3^x\cdot3^2+3^x\cdot3^3=360\\ 3^x\left(1+3+3^2+3^3\right)=360\\ 3^x\cdot40=360\\ 3^x=9\\ 3^x=3^2\\ =>x=2\)

\(\left(\dfrac{1}{2}+1\right)\cdot\left(\dfrac{1}{3}+1\right)\cdot\left(\dfrac{1}{4}+1\right)\cdot...\cdot\left(\dfrac{1}{99}+1\right)\\ =\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{100}{99}\\ =\dfrac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot.....\cdot99}\\ =\dfrac{100}{2}=50\)

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2};\dfrac{1}{3^2}< \dfrac{1}{2\cdot3};.....;\dfrac{1}{50^2}< \dfrac{1}{49\cdot50}\\ =>\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+....+\dfrac{1}{49\cdot50}=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{49}-\dfrac{1}{50}=\left(1+1\right)-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-.....-\left(\dfrac{1}{49}-\dfrac{1}{49}\right)-\dfrac{1}{50}=2-0-0-0.....-\dfrac{1}{50}\\ =2-\dfrac{1}{50}< 2=>\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 2\)

\(\dfrac{13}{8}+\dfrac{9}{24}+\left(\dfrac{27}{8}+\dfrac{15}{24}\right)\\=\left(\dfrac{13}{8}+\dfrac{27}{8}\right)+\left(\dfrac{9}{24}+\dfrac{15}{24}\right)\\ =\dfrac{40}{8}+\dfrac{24}{24}\\ =5+1\\ =6\)

Chu vi cái sân hcn là: \(\left(\dfrac{2}{3}+38,4\right)\cdot2=\dfrac{1172}{15}\left(m\right)\)

Diện tích cái sân hcn là:  \(\dfrac{2}{3}\cdot38,4=25,6\left(m^2\right)\)

+) Để \(\dfrac{n-5}{n-3}\) có giá trị nguyên thì \(n-5⋮n-3\)

 \(n-5⋮n-3\\ =>n-5-\left(n-3\right)⋮n-3\\ =>n-5-n+3⋮n-3\\ =>-2⋮n-3\\ =>n-3\inƯ\left(-2\right)\\ Ư\left(-2\right)=\left\{-1;1;-2;2\right\}\\ TH1:n-3=-1=>n=2\\ TH2:n-3=1=>n=4\\ TH3:n-3=-2=>n=1\\ TH4:n-3=2=>n=5\)

\(M=5+5^2+5^3+5^4...+5^{80}\\ M=5\left(1+5\right)+5^3\left(1+5\right)+5^5\left(1+5\right)+.....+5^{79}\left(1+5\right)\\ M=5\cdot6+5^3\cdot6+5^5\cdot6+....+5^{79}\cdot6\\ M=6\left(5+5^3+5^5+5^7+....+5^{79}\right)⋮6=>M⋮6\)