1/ 24.315+3.8.561+4.6.124

  =24.315+24.561+24.124

  =24.(315+561+124)

  =24.1000

  =24000

2/1+3+...+99-500

  Ta tính tổng của 1+3+...+99

SSH (99-1):2+1=50(số)

Tổng (99+1).50:2=50.50=2500

1+3+...+99-500=2500-500=2000

đáp án = 12 , cách giải hơi dài nên mik ko ghi dc

Bạn ghi lời giải được không, co mình bắt phải có lời giải

a,   \(\frac{24.315+3.561.8+4.124.6}{1+3+5+7+...+97+99-500}\) (1) 

Đặt : S = 1 + 3 + 5 + 7 + ... + 97 + 99 

SSH của S là : (99 -1) : 2 + 1 = 50(sh) 

Tổng của S là : \(\frac{\left(99+1\right).50}{2}=\frac{100.50}{2}=\frac{5000}{2}=2500\)

Thay S vào biểu thức (1) Ta có : 

\(\frac{24.315+3.561.8+4.124.6}{2500-500}\)

\(=\frac{3.8.315+3.561.8+4.2.124.3}{2000}\)

\(=\frac{3.8.315+3.561.8+8.124.3}{2000}\)

 \(=\frac{\left(3.8\right).\left(315+561+124\right)}{2000}=\frac{24.1000}{2000}=\frac{24000}{2000}=12\)

b, \(\frac{3^9.3^{20}.2^8}{3^{24}.243.2^6}=\frac{3^{29}.2^8}{3^{24}.3^5.2^6}=\frac{3^{29}.2^6.2^2}{3^{29}.2^6}=2^2=4\)

khó thể xem trên mạng

\(\dfrac{1}{500}+\dfrac{3}{500}+\dfrac{5}{500}+...+\dfrac{95}{500}+\dfrac{97}{500}+\dfrac{99}{500}\)

\(=\left(\dfrac{1}{500}+\dfrac{99}{500}\right)+\left(\dfrac{3}{500}+\dfrac{97}{500}\right)+\left(\dfrac{5}{500}+\dfrac{95}{500}\right)+...\)

\(=\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}+...\) ( 50 số )

\(=\dfrac{1}{5}.50\)

\(=10\)

Nguyễn Huy TúAce Legonasoyeon_Tiểubàng giảiTrần Việt Linh

Võ Đông Anh TuấnHoàng Lê Bảo NgọcPhương An

Đặt:

\(A=\dfrac{1}{500}+\dfrac{3}{500}+\dfrac{5}{500}+...+\dfrac{97}{500}+\dfrac{99}{500}\)

\(\Rightarrow A=\dfrac{1+3+5+...+97+99}{500}\)

\(\Rightarrow A=\dfrac{\dfrac{\left[\left(99-1\right):2+1\right].\left(99+1\right)}{2}}{500}=\dfrac{2500}{500}\)

\(\Rightarrow A=5\)

Chúc bạn học tốt!

Mình làm được một câu thôi, bạn dựa vào làm nha!

Ta có: \(A=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1\cdot99}+\dfrac{1}{3\cdot97}+\dfrac{1}{5\cdot95}+...+\dfrac{1}{97\cdot3}+\dfrac{1}{99\cdot1}}\)
\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{100}{1\cdot99}+\dfrac{100}{3\cdot97}+\dfrac{100}{5\cdot95}+...+\dfrac{100}{97\cdot3}+\dfrac{100}{99\cdot1}}\)

\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{1+\dfrac{1}{99}+\dfrac{1}{3}+\dfrac{1}{97}+\dfrac{1}{5}+\dfrac{1}{95}+...+\dfrac{1}{97}+\dfrac{1}{3}+\dfrac{1}{99}+1}\)

\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{2\left(1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}\right)}\)

\(\Leftrightarrow\dfrac{A}{100}=\dfrac{1}{2}\)

hay A=50

\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7} +.....................+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+....+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)

\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+..........+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+.......+\dfrac{1}{49.51}\right)}\)

\(=\dfrac{\dfrac{100}{1.99}+\dfrac{100}{3.97}+...........+\dfrac{100}{49.51}}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+...........+\dfrac{1}{49.51}\right)}\)

\(=\dfrac{100\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+.............+\dfrac{1}{49.51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+..........+\dfrac{1}{49.51}\right)}\)

\(=\dfrac{100}{2}\)

\(=50\)

\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+.....+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}}=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(\dfrac{1}{3}+\dfrac{1}{97}\right)+....+\left(\dfrac{1}{49}+\dfrac{1}{51}\right)}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+.....+\dfrac{1}{49.51}\right)}=\dfrac{\dfrac{100}{99}+\dfrac{100}{3.97}+....+\dfrac{100}{49.51}}{2\left(\dfrac{1}{1.99}+\dfrac{1}{3.97}+....+\dfrac{1}{49.51}\right)}=\dfrac{100}{2}=50\)