A=(1-1/3)x(1-1/4)x(1-1/5)x......x(1-1/99). Tim A
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27 tháng 5 2022
Bài 2:
a: \(\left(-1\right)\cdot\left(-2\right)\cdot\left(-3\right)\cdot\left(-4\right)\cdot\left(-5\right)\cdot\left[\left(-3\right)-\left(-5\right)\right]\)
\(=-\left(1\cdot2\cdot3\cdot4\cdot5\right)\cdot\left[-3+5\right]\)
\(=-120\cdot2=-240\)
b: \(1-2+3-4+5-6+...-98+99\)
=(-1)+(-1)+...+(-1)+99
=99-49=50
Q
22 tháng 9 2017
2a) (-1).(-2).(-3).(-4).(-5).[(-3)-(-5)]
= (-1).(-2).(-3).(-4).(-5).2
=> -240
2b) 1-2+3-4+5-6+...+98+99
= (1-2)+(3-4)+(5-6)+...+(97-98)+99
=> Ta có 98 cặp
= (-1)+(-1)+(-1)+...+(-1)+99
= 98(-1)+99
= (-98)+99
= 1
3a) (x-1)(y-2) = 5
=> x-1;y-2 \(\in\) Ư(5) = {-1,-5,1,5}
Ta có bảng :
| x-1 | -1 | -5 | 1 | 5 |
| y-2 | -5 | -1 | 5 | 1 |
| x | 0 | -4 | 2 | 6 |
| y | -3 | 1 | 7 | 3 |
Vậy x = {0,-4,2,6}
2b) x(y-3)=12
=> x;y-3 \(\in\) Ư(12) = {-1,-2,-3,-4,-12,1,2,4,12}
Tương đương với x = {-1,-2,-3,-4,-12,1,2,4,12}
16 tháng 8 2019
\(a,x-7\frac{5}{8}=1\frac{1}{4}\)
=> \(x-\frac{61}{8}=\frac{5}{4}\)
=> \(x=\frac{5}{4}+\frac{61}{8}\)
=> \(x=\frac{10}{8}+\frac{61}{8}=\frac{71}{8}=8\frac{7}{8}\)
\(b,x+7\frac{5}{8}=9\frac{1}{4}\)
=> \(x+\frac{43}{5}=\frac{37}{4}\)
=> \(x=\frac{37}{4}-\frac{43}{5}=\frac{13}{20}\)
\(c,\left[x-7\frac{5}{8}\right]:\frac{1}{2}=3\)
=> \(\left[x-\frac{61}{8}\right]=3\cdot\frac{1}{2}\)
=> \(\left[x-\frac{61}{8}\right]=\frac{3}{2}\)
=> \(x-\frac{61}{8}=\frac{3}{2}\)
=> \(x=\frac{3}{2}+\frac{61}{8}=\frac{12}{8}+\frac{61}{8}=\frac{73}{8}=9\frac{1}{8}\)
d, \(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+\frac{x}{5\cdot7}+...+\frac{x}{97\cdot99}=99\)
=> \(\frac{x}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right]=99\)
=> \(\frac{x}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=99\)
=> \(\frac{x}{2}\left[1-\frac{1}{99}\right]=99\)
=> \(\frac{x}{2}\cdot\frac{98}{99}=99\)
=> \(\frac{98x}{198}=99\)
=> 98x = 99 . 198
=> 98x = 19602
=> x = 19602 : 98 = 9801/49
16 tháng 8 2019
a) \(x-7\frac{5}{8}=1\frac{1}{4}\)
=> \(x=\frac{5}{4}+\frac{61}{8}\)
=> \(x=\frac{71}{8}\)
b) \(x+7\frac{5}{8}=9\frac{1}{4}\)
=> \(x=\frac{37}{4}-\frac{61}{8}\)
=> \(x=\frac{13}{8}\)
c) \(\left(x-7\frac{5}{8}\right):\frac{1}{2}=3\)
=> \(x-\frac{61}{8}=3.\frac{1}{2}\)
=> \(x-\frac{61}{8}=\frac{3}{2}\)
=> \(x=\frac{3}{2}+\frac{61}{8}\)
=> \(x=\frac{73}{8}\)
d) \(\frac{x}{1.3}+\frac{x}{3.5}+...+\frac{x}{97.99}=99\)
=> \(x.\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)=99\)
=> \(\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)=99\)
=> \(x\left(1-\frac{1}{99}\right)=99:\frac{1}{2}\)
=> \(x.\frac{98}{99}=198\)
=> \(x=198:\frac{98}{99}=\frac{9801}{49}\)
NG
1
2 tháng 3 2017
\(A=\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{5}\right)\cdot...\cdot\left(1-\frac{1}{99}\right)\)
\(A=\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{98}{99}\)
\(A=\frac{2}{99}\)
P
0
4 tháng 4 2016
A = 2/3 . 3/4 . 4/5 ..... 98/99
A = 2 . 3 . 4 ....... 98/3 . 4 . 5 ...... 99
A = 2/99
DT
5
2/99 nha bạn mình làm rồi
ở violympic đúng ko
\(\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{5}\right)\cdot...\cdot\left(1-\frac{1}{99}\right)\)
\(=\frac{2}{3^{\left(1\right)}}\cdot\frac{3^{\left(1\right)}}{4^{\left(1\right)}}\cdot\frac{4^{\left(1\right)}}{5^{\left(1\right)}}\cdot...\cdot\frac{98^{\left(1\right)}}{99}\)
\(=\frac{2}{99}\)