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\(15.95+5.95-\left(3^2\right)^3+5^7:5^4\)
\(=\left(15+5\right).95-3^6+5^2\)
\(=20.95-729-25\)
\(=1900-729-25\)
\(=1171-25\)
\(=1146\)
\(15.95+5.95-\left(3^2\right)^3+5^7:5^4\)
\(=\left(15+5\right)\cdot95-3^6+5^3\)
\(=20\cdot95-729-125\)
\(=1900-729-125\)
\(=1046\)
Goi bieu thuc tren la A
A= 52017: 52015+ 52016: 52015- 52015: 52015
A= 52017-2015+ 52016-2015- 52015-2015
A= 52+ 51+50
A= 25 +5 +1
A= 31
A
Ta có:j
\(\left(5^{2017}+5^{2016}-5^{2015}\right)\div5^{2015}=5^{2017}\div5^{2015}+5^{2016}\div5^{2015}-5^{2015}\div5^{2015}\)
\(=5^2+5-1=25+5-1=29\)
Vậy giá trị của biểu thức là 29
a) \(D=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\)
\(\Rightarrow7D=1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}\)
\(\Rightarrow7D-D=\left(1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{99}}\right)-\left(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{100}}\right)\)
\(\Rightarrow6D=1-\frac{1}{7^{100}}\)
\(\Rightarrow D=\left(1-\frac{1}{7^{100}}\right).\frac{1}{6}\)
\(S=5+5^1+5^2+5^3+...+5^{2024}\)
\(=5+\left(5^1+5^2+5^3+5^4\right)+\left(5^5+5^6+5^7+5^8\right)+...+\left(5^{2021}+5^{2022}+5^{2023}+5^{2024}\right)\)
\(=5+\left(5^1+5^2+5^3+5^4\right)+5^4\left(5^1+5^2+5^3+5^4\right)+...+5^{2020}\left(5^1+5^2+5^3+5^4\right)\)
\(=5+780\left(1+5^4+...+5^{2020}\right)\)
Có \(780⋮65\)nên \(780\left(1+5^4+...+5^{2020}\right)⋮65\)
suy ra \(S\)chia cho \(65\)dư \(5\).
TA CÓ:6A= 1.6+6.6+6.6^2+..........+6^1000.6
6A= 6+6^2+6^3+ +6^1000+6^1001
A=1+6+6^2+........+6^1000
6A-A=6^1001-1
vì 6^1001 chia hết cho 6:;1 chia 6 dư 5 suy ra A chia 6 dư 5