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\(\frac{x}{1.3}+\frac{x}{3.5}+\frac{x}{5.7}+....+\frac{x}{97.99}=\frac{49}{99}\)
\(\Leftrightarrow\frac{x}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99}\right)=\frac{49}{99}\)
\(\Leftrightarrow\frac{x}{2}.\left(\frac{1}{1}-\frac{1}{99}\right)=\frac{49}{99}\)
\(\Leftrightarrow\frac{x}{2}.\frac{98}{99}=\frac{49}{99}\)
\(\Leftrightarrow\frac{x}{2}=\frac{49}{99}\div\frac{98}{99}\)
\(\Leftrightarrow\frac{x}{2}=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{2}\times2=1\)
\(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+...+\frac{x}{97\cdot99}=\frac{49}{99}\)
\(\Rightarrow\frac{x}{2}\left[\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{97\cdot99}\right]=\frac{49}{99}\)
\(\Rightarrow\frac{x}{2}\left[\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=\frac{97}{99}\)
\(\Rightarrow\frac{x}{2}\left[1-\frac{1}{99}\right]=\frac{49}{99}\)
\(\Rightarrow\frac{x}{2}\cdot\frac{98}{99}=\frac{49}{99}\)
\(\Rightarrow\frac{x}{2}=\frac{1}{2}\)
=> x = 1/2 * 2 = 1
\(A=1.3+3.5+5.7+...+45.47+47.49\)
\(A=\left(1.49\right)+\left(2.3\right)+\left(2.5\right)+\left(2.7\right)+.....+\left(2.47\right)\)
\(A=49+2.\left(3+5+7+....+47\right)\)
Bây giờ ta phải tìm SSH của :
\(3+7+...+47\)
Vậy SSH của tổng đó là :
(47-3):2+1=23 (SSH)
=> \(A=49+2.\left(\frac{\left(47+3\right).23}{2}\right)\)
\(A=49+2.575\)
\(A=49+1150\)
\(A=1199\)
Dạng này lầm đầu gặp ![]()
A=1.3+3.5+5.7+...+95.97+97.99
6A=1.3.6+3.5.6+5.7.6+...+95.97.96+97.99.96
=1.3.(5+1)+3.5.(7-1)+...+95.97.(99-93)+97.99.(101-95)
=1.1.3+1.3.5-1.3.5+3.5.7-....-95.97.99+97.99.101
=3.97.99.101
=>A=\(\frac{3+97.99.101}{6}=\frac{1+97.33.101}{2}\)\(=161651\)
a)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{99}-\frac{1}{101}\right)\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
b) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}=\frac{2}{1.3}.\frac{5}{2}+\frac{2}{3.5}.\frac{5}{2}+\frac{2}{5.7}.\frac{5}{2}+...+\frac{2}{99.101}.\frac{5}{2}\)
\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)
a.2/1.3+2/3.5+2/5.7+................+2/99.101
1-1/3+1/3-1/5+1/5-1/7+....+1/99-1/101
1-1/101
100/101
b.5/1.3+5/3.5+5/5.7+............+5/99.101
5.2/1.3.2+5.2/3.5.2+5.2/5.7.2+........+5.2+99.101.2
5/2(2/1.3+2/3.5+2/5.7+........+2/99.101)
5/2(1-1/3+1/3-1/5+1/5-1/7+........+1/99-1/101)
5/2(1-1/101)
5/2.100/101
250/101
2S=2/1.3+2/3.5+....+2/99.101
2S=1-1/3+1/3-1/5+....+1/99-1/101
2S=1-1/101
2S+1/101=1-1/101+1/101=1
Nho tick nha
\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(S=1-\frac{1}{101}=\frac{100}{101}\)
\(2S+\frac{1}{101}=\frac{100}{101}\)
\(S=2.\frac{100}{101}+\frac{1}{101}\)
\(\Rightarrow S=\frac{201}{101}\)
****
Ta có: A = 1.3 + 3.5 + 5.7 +…+ 97.99 + 99.101
A = 1.(1 + 2) + 3.(3 + 2) + 5.(5 + 2) + … + 97.(97 + 2) + 99.(99 + 2)
A = (12 + 32 + 52 + … + 972 + 992) + 2.(1 + 3 + 5 + … + 97 + 99).
Đặt B = 12 + 32 + 52 + … + 992
=> B = (12 + 22 + 32 + 42 + … + 1002) – 22.(12 + 22 + 32 + 42 + … + 502)
Áp dụng vào B ta được:
B = 100.101.201 : 6 – 4.50.51.101 : 6 = 166650
=> A = 166650 + 2.(1 + 99).50 : 2
=> A = 166650 + 5000 = 172650.
Đ/s: A = 172650.
de qua
Ta có: A = 1.3 + 3.5 + 5.7 +…+ 97.99 + 99.101
A = 1.(1 + 2) + 3.(3 + 2) + 5.(5 + 2) + … + 97.(97 + 2) + 99.(99 + 2)
A = (12 + 32 + 52 + … + 972 + 992) + 2.(1 + 3 + 5 + … + 97 + 99).
Đặt B = 12 + 32 + 52 + … + 992
=> B = (12 + 22 + 32 + 42 + … + 1002) – 22.(12 + 22 + 32 + 42 + … + 502)
Tính dãy tổng quát C = 12 + 22 + 32 + … + n2
C = 1.(0 + 1) + 2.(1 + 1) + 3.(2 + 1) + … + n.[(n – 1) + 1]
C = [1.2 + 2.3 + … + (n – 1).n] + (1 + 2 + 3 + … + n)
C = = n.(n + 1).[(n – 1) : 3 + 1 : 2] = n.(n + 1).(2n + 1) : 6
Áp dụng vào B ta được:
B = 100.101.201 : 6 – 4.50.51.101 : 6 = 166650
=> A = 166650 + 2.(1 + 99).50 : 2
=> A = 166650 + 5000 = 172650.
Đ/s: A = 172650.
Chúc bn học tốt
=^.^=