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S= 1/2.4+1/4.6+1/6.8+1/8.10
S= 1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10
S= 1/2-1/10
S= 2/5
Sai thì bình luận ch mình biết nha
A=1/2.4+1/4.6+........+1/100.102
A=1/2-1/4+1/4-1/6+.......+1/100-1/102
A=1/2-1/102
A=51/102-1/102
A=50/102
A=25/51
\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{18}-\frac{1}{20}\)
\(A=\frac{1}{2}-\frac{1}{20}\)
\(A=\frac{10}{20}-\frac{1}{20}\)
\(A=\frac{9}{20}\)
Số hạng thứ 50 của dãy là: \(\frac{1}{100.102}\)
Tổng 50 số hạng đầu của dãy là:\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+.....+\frac{1}{100.102}=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{100}-\frac{1}{102}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{102}\right)=\frac{1}{2}.\frac{25}{51}=\frac{25}{102}\)
phân số thứ 50 là 1/98.100
1/2.4+1/4.6+1/6.8+.......+1/98.100
=2.(1/2-1/4+1/4-1/6+1/6-1/8+.........+1/98-1/100).1/2
=(1-1/2+1/2-1/3+1/3-1/4+...........+1/49-1/50).1/2
=(1-1/50).1/2
=49/50.1/2
=49/100
\(S=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2018.2020}\)
\(S=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2018.2020}\right)\)
\(S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2018}-\frac{1}{2020}\right)\)
\(S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2020}\right)\)
Tự tính
S=1/2.4+1/4.6+1/6.8+...+1/2018.2020
S=1/2.(2/2.4+2/4.6+2/6.8+...+2/2018.2020)
S=1/2.(1-1/4+1/4-1/6+1/6-1/8+...+1/2018-1/2020)
S=1/2.(1-1/2020)
S=1/2.(2020/2020-1/2020)
S=1/2.2019/2020
S=2019/4040
Ta có : D = \(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+.....+\frac{4}{2008.2010}\)
\(\Leftrightarrow D=2\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{2008.2010}\right)\)
\(\Leftrightarrow D=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(\Leftrightarrow D=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(\Leftrightarrow D=1-\frac{1}{1005}=\frac{1004}{1005}\)
D = 2.(2/2.4+2/4.6+...+2/2008.2010)
=2(1/2-1/4+1/4-1/6+......+1/2008-1/2
=2(1/2-1/2010)
=2.502/1005
=1004/1005
A=3n+1/n-1=3(n-1)+4/n-1=3+4/n-1
Để A là số nguyên thì 4/n-1 là số nguyên
=>n-1 thuộc Ư(4)=1,-1,2,-2,4,-4
=>n thuộc (2,0,3,-1,5,-3)
Ta có : \(A=\frac{3n+2}{n-1}+\frac{3n-3+5}{n-1}=\frac{3\left(n-1\right)+5}{n-1}=\frac{3\left(n-1\right)}{n-1}+\frac{5}{n-1}=3+\frac{5}{n-1}\)
Để A có giá trị nguyên thì n - 1 thuộc Ư(5) = {-1;-5;1;5}
n - 1 | -5 | -1 | 1 | 5 |
n | -4 | 0 | 2 | 6 |
A = \(3+\frac{5}{n-1}\) | 2 | -2 | 8 | 4 |
\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)
\(S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\)
\(S=\frac{1}{2}-\frac{1}{10}\)
\(S=\frac{2}{5}\)
\(A=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{2022\cdot2024}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{2022\cdot2024}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2022}-\dfrac{1}{2024}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2024}\right)=\dfrac{1}{2}\cdot\dfrac{1011}{2024}=\dfrac{1011}{4048}\)
\(A=\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{2022.2024}\)
\(A=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2022}-\dfrac{1}{2024}\)
\(A=\dfrac{1}{2}-\dfrac{1}{2024}\)
\(A=\dfrac{1012}{2024}-\dfrac{1}{2024}\)
\(A=\dfrac{1211}{2024}\)
Vậy \(A=\dfrac{1211}{2024}\)