\(S=1^2+2^2+3^2+...+99^2\)
\(=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+99\left(100-1\right)\)
\(=\left(1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\right)-\left(1+2+3+...+99\right)\)
\(=\frac{99\cdot100\cdot101}{3}-\frac{99\cdot\left(99+1\right)}{2}\)
\(=333300-4950\)
\(=328350\)

\(M=1\cdot3+3\cdot5+5\cdot7+...+97\cdot99\)
\(=3+\frac{3\cdot5\cdot\left(7-1\right)+5\cdot7\cdot\left(9-3\right)+...+97\cdot99\cdot\left(101-95\right)}{6}\)
\(=3+\frac{3\cdot5\cdot7-1\cdot3\cdot5+5\cdot7\cdot9-3\cdot5\cdot7+...+97\cdot99\cdot101-95\cdot97\cdot99}{6}\)
\(=3+\frac{-\left(1\cdot3\cdot5\right)}{6}+\frac{3\cdot5\cdot7+5\cdot7\cdot9-3\cdot5\cdot7+...+97\cdot99\cdot101-95\cdot97\cdot99}{6}\)
\(=3+-\frac{15}{6}+\frac{97\cdot99\cdot101}{6}\)
\(=3+-2,5+161650,5\)
\(=161651\)

Câu a:

2\(^{300}\) và 3\(^{200}\)

2\(^{300}\) = (2\(^3\))\(^{100}\) = 8\(^{100}\)

3\(^{200}\) = (3\(^2\))\(^{100}\) = 9\(^{100}\)

8\(^{100}\) < 9\(^{100}\)

Vậy 2\(^{300}\) < 3\(^{200}\)

câu b:

99\(^{20}\) và 9999\(^{10}\)

99\(^{20}\) = (99\(^2\))\(^{10}\) = 9801\(^{10}\)

9999\(^{10}\) > 9801\(^{10}\)

Vậy 99\(^{20}\) < 9999\(^{10}\)

Câu c:

3\(^{500}\) và \(7^{300}\)

3\(^{500}\) = (3\(^5\))\(^{100}\) = 243\(^{100}\)

7\(^{300}\) = (7\(^3\))\(^{100}\) = 343\(^{100}\)

243\(^{100}\) < 343\(^{100}\)

Vậy 3\(^{500}\) < 7\(^{300}\)

Câu d:

11\(^{1979}\) và 37\(^{1320}\)

11\(^{1979}\) < 11\(^{1980}\) = (11\(^3\))\(^{660}\) = 1331\(^{660}\)

37\(^{1320}\) = (37\(^2\))\(^{660}\) = 1369\(^{660}\)

1331\(^{660}<1369^{660}\)

Vậy 11\(^{1979}\) < 37\(^{1320}\)

a, 1 - 7x = 3x - 4

=> -7x - 3x = - 4 - 1

=> - 10x = - 5

=> x = 1/2

vậy_

b, đặt  \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

\(3A-A=1-\frac{1}{3^{99}}\)

\(A=\frac{1-\frac{1}{3^{99}}}{2}\)

mk chỉ bt lm mấy phần hui à!

d)\(\frac{5}{17}+\frac{-4}{7}-\frac{20}{31}+\frac{12}{17}-\frac{11}{31}\)\(=\left(\frac{5}{17}+\frac{12}{17}\right)+\left(\frac{-20}{31}-\frac{11}{31}\right)+\frac{-4}{7}\)

\(=\frac{17}{17}+\frac{-31}{31}+\frac{-4}{7}\)\(=1+\left(-1\right)+\frac{-4}{7}\)\(=0+\frac{-4}{7}\)\(=-\frac{4}{7}\)

e)\(\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{20}{7}-\frac{13}{3}+\frac{13}{23}}\)

Đặt \(A=\frac15-\frac{1}{5^3}+\frac{1}{5^5}-\frac{1}{5^7}+\cdots-\frac{1}{5^{99}}\)

=>\(25A=5-\frac15+\frac{1}{5^3}-\frac{1}{5^5}+\cdots-\frac{1}{5^{97}}\)

=>\(A+25A=\frac15-\frac{1}{5^3}+\frac{1}{5^5}-\frac{1}{5^7}+\cdots-\frac{1}{5^{99}}+5-\frac15+\frac{1}{5^3}-\frac{1}{5^5}+\cdots-\frac{1}{5^{97}}\)

=>\(26A=5-\frac{1}{5^{99}}=\frac{5^{100}-1}{5^{99}}\)

=>\(A=\frac{5^{100}-1}{5^{99}\cdot26}\)

rút gọn nx :)

Đặt \(A=\frac15-\frac{1}{5^3}+\frac{1}{5^5}-\frac{1}{5^7}+\cdots-\frac{1}{5^{99}}\)

=>\(25A=5-\frac15+\frac{1}{5^3}-\frac{1}{5^5}+\cdots-\frac{1}{5^{97}}\)

=>\(A+25A=\frac15-\frac{1}{5^3}+\frac{1}{5^5}-\frac{1}{5^7}+\cdots-\frac{1}{5^{99}}+5-\frac15+\frac{1}{5^3}-\frac{1}{5^5}+\cdots-\frac{1}{5^{97}}\)

=>\(26A=5-\frac{1}{5^{99}}=\frac{5^{100}-1}{5^{99}}\)

=>\(A=\frac{5^{100}-1}{5^{99}\cdot26}\)

câu thứ 2 =0 vì (63.1,-21.3,6)=0

MIK muốn hỏi câu đầu tiên

a) \(2007^{2008}=\left(2007^4\right)^{502}\)

\(=\left(...1\right)^{502}=\left(...1\right)\)

=> \(2007^{2008}\) có chữ số tận cùng là 1

b) \(1358^{2009}=\left(1358^4\right)^{502}\cdot1358\)

\(=\left(...6\right)^{502}\cdot1358=\left(...6\right)\cdot1358=\left(...8\right)\)

=> \(1358^{2009}\) có chứ số tận cùng là 8

c) \(52^{35}=\left(52^4\right)^8\cdot52^3\)

\(=\left(...6\right)^8\cdot\left(...8\right)=\left(...6\right)\cdot\left(...8\right)=\left(...8\right)\)

=> \(52^{35}\) có chữ số tận cùng là 8

d) \(9^{99}=\left(9^2\right)^{49}\cdot9\)

\(=\left(...1\right)^{49}\cdot9=\left(...9\right)\)

=> \(9^{99}\) có chữ số tận cùng là 9

e) \(5^{6^7}\) có chữ số tận cùng bằng 5 là số lẻ

\(\Rightarrow5^{6^7}=2k+1\) ( \(k\in N\)* )

\(\Rightarrow4^{5^{6^7}}=4^{2k+1}=16^k\cdot4\)

\(=\left(...6\right)\cdot4=\left(...4\right)\)

\(\Rightarrow\text{ 4}^{5^{6^7}}\) có chữ số tận cùng là 4

g) \(=\left(17^4\right)^{503}+\left(...1\right)-\left(7^4\right)^{503}\)

\(=\left(...1\right)^{503}+\left(...1\right)-\left(...1\right)^{503}\)

\(=\left(...1\right)+\left(...1\right)-\left(...1\right)=\left(...1\right)\)có tận cùng là 1

h) \(=\left(3^4\right)^{505}\cdot3\cdot\left(7^4\right)^{505}\cdot7^2\cdot\left(13^4\right)^{505}\cdot13^3\)

\(=81^{505}\cdot3\cdot\left(...1\right)^{505}\cdot49\cdot\left(...1\right)^{505}\cdot\left(...7\right)\)

\(=\left(...1\right)\cdot3\cdot\left(...1\right)\cdot49\cdot\left(...1\right)\cdot\left(...7\right)\)

\(=\left(...9\right)\) có chữ số tận cùng là 9

Sao mà bạn học đến Chương III rồi vậy, học thêm à