[1x1]x....x[99x100]=?
Những câu hỏi liên quan
a] A=1x2 +2x3+3x4+...+ 99x100
b]B=1x100+2x99+3x98+99x2+100x1
c] C=1x1+2x2+3x3+...+100x100
A = (100 - 1x1) x (100 - 2x2) x....x(100-10x10) =?
[1+1/2]x[1+1/3]x[1+1/4]x..................x[1x1/2010]
\(\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot...\cdot\left(1+\frac{1}{2010}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{2011}{2010}\)
\(=\frac{2011}{2}\)
\(=1005,5\)
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P = (1-2/2x3)x(1-2/3x4)x(1-2/4x5) . . . x(1-2/99x100)
\(P=\left(1-\frac{2}{2.3}\right).\left(1-\frac{2}{3.4}\right).\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)\)
\(P=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}...\frac{9898}{99.100}\)
\(P=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{98.101}{99.100}\)
\(P=\frac{1.2.3...98}{2.3.4...99}.\frac{4.5.6...101}{3.4.5...100}\)
\(P=\frac{1}{99}.\frac{101}{3}=\frac{101}{297}\)
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tinh A =(1+1/1x3) x(1+1/2x4)x(1+1/3x5)x...(1+1/99x100)
Ps cuối hình như có vấn đề..........
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60x(7/12+4/15)
(1-1/2)x(1-1/3)x(1x1/4)x(1/5)
a: \(=60\cdot\dfrac{17}{20}=51\)
b: \(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}=\dfrac{1}{5}\)
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Giới hạn lim x → 0 - 1 x 1 x + 1 - 1 bằng
A. 0
B. -1
C. 1
D. - ∞
1 x 1x 1x 1x1 x1 1x1x 1x 1 =