1,(1/2 +1).(1/3 +1).(1/4 +1)....(1/999 +1)
2,(1/2 -1).(1/3 -1).(1/4 -1)....(1/1000 -1)
3, 3/22 . 8/32 . 15/42 ..... 99/102
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Khách
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Những câu hỏi liên quan
10 tháng 3 2017
a) =3/2*4/3*5/4*....* 1000/999
=3*4*5*......*1000 / 2*3*4*...*999
=1000/2=500
phần b;c mk chưa làm đc
HD
28 tháng 3 2017
b) ...........
= 1/2x2/3x.....x999/1000
= 1x2x...x999/2x3x...x1000
=1/1000
HD
28 tháng 3 2017
a) ...............
= 3/2 . 4/3 .... 1000/999
= 3x4x5x....x1000/2x3x4x...x999
=1000/2=500
15 tháng 3 2016
a. =3/2*4/3*5/4*...*1000/999
=3*4*5*....*1000/2*3*4*...*999
=1000/2
=500
b. =-1/2*-2/3*-3/4*....*-999/1000
=-1*-2*-3*.....*-999/2*3*4*....*1000
=-1/1000
H9
HT.Phong (9A5)
CTVHS
22 tháng 6 2023
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
29 tháng 3 2017
a,\(\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{999}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}.....\dfrac{1000}{999}\)
\(=\dfrac{3.4.5....1000}{2.3.4....999}=\dfrac{1000}{2}=500\)
b,\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{1000}-1\right)\)
\(=\dfrac{-1}{2}.\dfrac{-2}{3}.\dfrac{-3}{4}.....\dfrac{-999}{1000}\)
=\(\dfrac{-\left(1.2.3....999\right)}{2.3.4....1000}=\dfrac{-1}{1000}\)
c,\(\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}....\dfrac{99}{10^2}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}....\dfrac{9.11}{10.10}\)
\(=\dfrac{1.3.2.4.3.5....9.11}{2.2.3.3.4.4....10.10}\)
\(=\dfrac{1.2.3...9}{2.3.4...10}.\dfrac{3.4.5...11}{2.3.4...10}\)
\(=\dfrac{1}{10}.\dfrac{11}{2}=\dfrac{11}{20}\)
21 tháng 3 2016
Vậy xét là \(\frac{1}{2}+1\)nhé.
a,\(\frac{3}{2}x\frac{4}{3}x\frac{5}{4}x...x\frac{1000}{999}\)
=3x4x5x...x1000/2x3x4x...x999
=1000/2=500
b, c tương tự câu a
21 tháng 3 2016
)(1/2+1)x(1/3+1)x(1/4+1)x...x(1/999+1)
b)(1/2-1)x(1/3-1)x(1/4-1)x...x(1/1000-1)
c)3/22 x 8/32 x 15/42 x .... x 99/102
mình ko biết làm chép lại de thui
Q
10 tháng 4 2017
a, \(\left(\dfrac{1}{2}+1\right).\left(\dfrac{1}{3}+1\right).\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{999}+1\right)\)
\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{1000}{999}\)
\(=\dfrac{3.4.5...1000}{2.3.4...999}\)
\(=\dfrac{1000}{2}\)\(=500\)
Q
10 tháng 4 2017
b, \(\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{1000}-1\right)\)
\(=\dfrac{-1}{2}.\dfrac{-2}{3}.\dfrac{-3}{4}...\dfrac{-999}{1000}\)
\(=\dfrac{\left(-1\right).\left(-2\right).\left(-3\right)...\left(-999\right)}{2.3.4...1000}\)
\(=\dfrac{-1}{1000}\)
