1/1*2+1/2*3+1/3*4+.......+1/999*1000+1
Những câu hỏi liên quan
A=1/1×2+1/3×4+1/4×5+...1/999×1000
B=1/501×1000+1/502×999+...+1/999×502+1/1000×501
Tính A/B
Tính nhanh : \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt[1]{2}+\sqrt[2]{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt[3]{4}+\sqrt[4]{5}}+...+\frac{1}{\sqrt{999}+\sqrt{1000}}+\frac{1}{\sqrt[999]{1000}+\sqrt[1000]{1001}}\)
1-1/2+1/3-1/4+......+1/999-1/1000
500-500/501-501/502-502/503-....-999/1000
các bạn ơi giúp nhanh nha mình đang cần rất gấp
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tính B=(2016/1000+2016/999+2016/998+...+2016/501)/(-1/1*2+/-1/3*4+-1/5*6+...+-1/999*1000)
\(B=\frac{\frac{2016}{1000}+\frac{2016}{999}+...+\frac{2016}{501}}{\frac{-1}{1.2}+\frac{-1}{3.4}+...+\frac{-1}{999.1000}}=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{999.1000}\right)}\)
\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}\right)}\)
\(=\frac{2016\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{999}+\frac{1}{1000}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\right)\right]}\)
\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{999}+\frac{1}{1000}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{500}\right)\right]}\)
\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+....+\frac{1}{999}+\frac{1}{1000}\right)}=\frac{2016}{-1}=-2016\)
Vậy B = - 2016
Bạn Xyz cho mik hỏi ở phần mẫu số tại sao lại có -2*(1/2+1/4+...+1/1000) vậy? Nó ở đâu ra thế?
1/1*2*3+1/2*3*4+1/3*4*5+.........+1/998*999*1000
1/1*2+1/2*3+1/3*4+...+1/999*1000
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{999\cdot1000}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}\)
\(=1-\frac{1}{1000}\)
\(=\frac{999}{1000}\)
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1/1*2+1/2*3+1/3*4+...+1/999*1000
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}\)
\(=1-\frac{1}{1000}\)
\(=\frac{999}{1000}\)
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1/1*2 + 1/2*3 + 1/3*4 + ... + 1/999*1000
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}\)
\(=1-\frac{1}{1000}=\frac{999}{1000}\)
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1/(1×2) + 1/ ( 2×3) +1/ (3×4) +................+1/(999×1000) +1
Xem thêm câu trả lời
1/1*2+1/2*3+1/3*4+...+1/999*1000+1=?
1/1*2+1/2*3+,,,,,+1/999*1000+1
=1/1-1/2+1/2-1/3+,,,,+1/999-1/1000+1
=1-1/1000+1
=1+1-1/1000
=2-1/1000
=1999/1000
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\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}+1\)
Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}\)
\(\Rightarrow A=1-\frac{1}{1000}=\frac{999}{1000}\)
Thay vào ta có : \(\frac{999}{1000}+1=\frac{1999}{1000}\)
Vậy ...
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1 / 1x2 + 1 / 2x3 + 1/ 3x4 +....+ 1/999x1000 + 1
= ( 1 / 1 x2 + 1/ 2 x3 + 1/ 3x4 + ....+ 1/999x 1000 ) + 1
= ( 1 x1 / 1 x2 + 1x2 / 2x3 + 1x 3 / 3x4 + ..... 1x 999 / 999x 1000 ) + 1
= ( 1 - 1 / 2 + 1 / 2 - 1 / 3 - 1 / 3 - 1 / 4 + ........+ 1 / 999 - 1 / 1000 ) + 1
= ( 1 - 1 / 1000 ) + 1
= 999 / 1000 + 1
= 1999 / 1000
Tk tớ nha
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