A=(1-1/2)×(1-1/3)×(1-1/4)×(1-1/5)×(1-1/2016)×(1-1/2017)
Những câu hỏi liên quan
A= ( 1/2017+ 2/2016+ 3/2015+...+ 2015/3+ 2016/2+ 2017) : ( 1/2+1/3+1/4+...+1/2017+1/2018)
Tính : A = ( 1 - 1/2 ) *( 1 - 1/3 ) *( 1 - 1/4 ) *( 1 - 1/5 ) ..... ( 1 - 1/2016 ) *( 1 - 1/2017 )
\(A=\left(1-\frac{1}{2}\right).\left(1-\frac{2}{3}\right)...\left(1-\frac{1}{2017}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}...\frac{2015}{2016}.\frac{2016}{2017}\)
\(A=\frac{1.2.3...2016}{2.3.4...2017}=\frac{1}{2017}\)
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\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{5}\right)\cdot...\cdot\left(1-\frac{1}{2016}\right)\cdot\left(1-\frac{1}{2017}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{2015}{2016}\cdot\frac{2016}{2017}\)
\(A=\frac{1\cdot2\cdot3\cdot4\cdot....\cdot2015\cdot2016}{2\cdot3\cdot4\cdot5\cdot....\cdot2016\cdot2017}\)
\(A=\frac{1}{2017}\)
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xin lỗi nha mình 0 có ngoặc tròn
A=[1-1/2]x[1-1/3]x...x[1-1/2017]
A=1/2x2/3x...x2016/2017
A=1-2016/2017
1/2017
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Bài 1 : Sắp xếp
5/4, 1/4, 1/23/50, 1/5, 4/5, 2006/2005, 2016/2017, 1/20, 3/55, 2017/2016, 3/58, 3/31, 1/12, 1/6, 3/5, 2/3, 3/52, 3/4, 1/15
Cho biểu thức A = 1/1*2*3+1/2*3*4+1/3*4*5+...+1/2015*2016*2017
Mk ko bt t mình nhé mk mới giam gia thôi
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\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2015.2016.2017}\)
\(A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{2017-2015}{2015.2016.2017}\)
\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2015.2016}-\frac{1}{2016.2017}\)
\(2A=\frac{1}{1.2}-\frac{1}{2016.2017}\)
\(A=\left(\frac{1}{1.2}-\frac{1}{2016.2017}\right)\div2\)
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\(2A=\frac{2^{2016}-4}{2^{2016}+1}=\frac{2^{2016}+1-5}{2^{2016}+1}=1-\frac{5}{2^{2016}+1}\)
\(2B=\frac{2^{2017}-4}{2^{2017}+1}=1-\frac{5}{2^{2017}+1}\)
Vì \(2^{2016} +1< 2^{2017}+1\)
=> \(2A< 2B\) Hay \(A< B\)
Chứng minh rằng F= 1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+4+5+6+...+2017)<2016/2017
Tính:
A=(1/1009+1/1010+...+1/2016+1/2017):(1-1/2+1/3-1/4+...+1/2018-1/2016+1/2017)
A = (1 - 1/2) x (1 - 1/3) x (1 - 1/4) x (1 - 1/5) x ... (1 - 1/2016) x (1 - 1/2017)
\(A=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times...\times\frac{2015}{2016}\times\frac{2016}{2017}=\frac{1}{2017}\)
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A = 1/2 + 1/3 +1/4 +.....+1/2016 + 1/2017 B = 2016/1 + 2015/2 + ......+ 2/2015 + 1/2016 . Tính B/A
\(\frac{B}{A}=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2016}+\frac{1}{2017}}\)
\(\frac{B}{A}=\frac{\left(\frac{2016}{1}+1\right)+\left(\frac{2015}{2}+1\right)+...+\left(\frac{1}{2016}+1\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(\frac{B}{A}=\frac{\frac{2017}{1}+\frac{2017}{2}+...+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(\frac{B}{A}=\frac{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}=2017\div\frac{1}{2017}=4068289\)
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