A= 1/1+2 + 1/1+2+3 + 1/1+2+3+4 + .... + 1/1+2+3+...+99
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PA
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Những câu hỏi liên quan
T
2
30 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`(1/2-1/3+1/4-1/5):(1/4-1/5)`
`=`\(\left(\dfrac{1}{6}+\dfrac{1}{20}\right)\div\dfrac{1}{20}\)
`=`\(\dfrac{1}{20}\div\dfrac{1}{20}+\dfrac{1}{6}\div\dfrac{1}{20}\)
`= 1+10/3`
`= 13/3`
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
30 tháng 6 2023
A = (\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)): (\(\dfrac{1}{4}\) - \(\dfrac{1}{5}\))
A = ( \(\dfrac{30}{60}\) - \(\dfrac{20}{60}\) + \(\dfrac{15}{60}\) - \(\dfrac{12}{60}\)):(\(\dfrac{5}{20}\) - \(\dfrac{4}{20}\))
A = \(\dfrac{13}{60}\): \(\dfrac{1}{20}\)
A = \(\dfrac{13}{60}\times\dfrac{20}{1}\)
A = \(\dfrac{13}{3}\)
24 tháng 2 2023
2A=1-1/2+1/2^2-...+1/2^98-1/2^99
=>3A=1-1/2^100
=>\(A=\dfrac{2^{100}-1}{3\cdot2^{100}}\)
LN
1
S
10 tháng 7 2018
\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+...+16\right)\)
\(=1+\frac{1}{2}\cdot\frac{2.3}{2}+\frac{1}{3}\cdot\frac{3.4}{2}+...+\frac{1}{16}\cdot\frac{16.17}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}\)
\(=\frac{2}{2}+\frac{3}{2}+...+\frac{17}{2}=\frac{1}{2}\left(2+3+...+17\right)=\frac{1}{2}\cdot\frac{16.19}{2}=4.19=76\)
15 tháng 1 2017
A =\(\left(1+\frac{2}{1}\right)\left(1+\frac{2}{2}\right)\left(1+\frac{2}{3}\right)\left(1+\frac{2}{4}\right)...\left(1+\frac{2}{26}\right)\left(1+\frac{2}{27}\right)\)
\(=\frac{3}{1}.\frac{4}{2}.\frac{5}{3}.\frac{6}{4}....\frac{28}{26}.\frac{29}{27}=\frac{28.29}{1.2}=14.29=406\)
TX
1
12 tháng 5 2023
a: \(A=\dfrac{2}{15}+\dfrac{13}{15}-\dfrac{1}{4}-\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{1}{2}\)
b: =5,4(-3,6-6,4)
=5,4*(-10)
=-54
TM
26 tháng 1 2022
1. ĐKXĐ: \(x\ne\pm1\)
2. \(A=\left(\dfrac{x+1}{x-1}-\dfrac{x+3}{x+1}\right)\cdot\dfrac{x+1}{2}\)
\(=\dfrac{\left(x+1\right)^2-\left(x-3\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)
\(=\dfrac{x^2+2x+1-x^2+4x-3}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)
\(=\dfrac{6x-2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)
\(=\dfrac{2\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x-3}{x-1}\)
3. Tại x = 5, A có giá trị là:
\(\dfrac{5-3}{5-1}=\dfrac{1}{2}\)
4. \(A=\dfrac{x-3}{x-1}\) \(=\dfrac{x-1-3}{x-1}=1-\dfrac{3}{x-1}\)
Để A nguyên => \(3⋮\left(x-1\right)\) hay \(\left(x-1\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\left(tmđk\right)\\x=0\left(tmđk\right)\\x=4\left(tmđk\right)\\x=-2\left(tmđk\right)\end{matrix}\right.\)
Vậy: A nguyên khi \(x=\left\{2;0;4;-2\right\}\)
\(A=\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{4950}=\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{9900}=2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=2.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=2.\left(\dfrac{1}{2}-\dfrac{1}{100}\right)=2.\dfrac{49}{100}=\dfrac{49}{50}\)