P=(1-1/2).(1-1/3).(1-1/4).(1-1/5)....(1-1/2011).(1-1/2012)
Những câu hỏi liên quan
1) Tìm x, y, z biết rằng x^2+y^2+z^2=xy+yz+xz và x^2011+y^2011+z^2011=3^2012
2) Tính A= (1^4+1/4)(3^4+1/4)(5^4+1/4)....(2011^4+1/4) / (2^4+1/4)(4^4+1/4)(6^4+1/4)....(2012^4+1/4)
x2+y2+z2= xy+yz+zx.
=> 2x2+2y2+2z2-2xy-2yz-2zx=0
=> ( x-y)2+(y-z.)2+(z-x)2 =0
=> x=y=z=0
Thay x=y=z vào x2011+y2011+z2011=32012 ta được:
3.x2011=3.32011
=> x2011=32011
=> x=3 hoặc x = -3
Hay x=y=z=3 hoặc x=y=z=-3
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1) có bn giải rồi ko giải nữa
2) \(A=\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)....\left(2011^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)....\left(2012^4+\frac{1}{4}\right)}\)
Với mọi n thuộc N ta có :
\(n^4+\frac{1}{4}=\left(n^4+2.\frac{1}{2}.n^2+\frac{1}{4}\right)-n^2=\left(n^2+\frac{1}{2}\right)^2-n^2=\left(n^2-n+\frac{1}{2}\right)\left(n^2+n+\frac{1}{2}\right)\)
\(=\left[n\left(n-1\right)+\frac{1}{2}\right]\left[n\left(n+1\right)+\frac{1}{2}\right]\)
Áp dụng ta được :
\(A=\frac{\frac{1}{2}\left(1.2+\frac{1}{2}\right)\left(2.3+\frac{1}{2}\right)\left(3.4+\frac{1}{2}\right)....\left(2011.2012+\frac{1}{2}\right)}{\left(1.2+\frac{1}{2}\right)\left(2.3+\frac{1}{2}\right)\left(3.4+\frac{1}{2}\right).......\left(2012.2013+\frac{1}{2}\right)}\)
\(=\frac{\frac{1}{2}}{2012.2013+\frac{1}{2}}=\frac{1}{8100313}\)
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1-1/2+1/3-1/4+1/5-1/6+...+1/2011-1/2012 / 1006-1006/1007-1007/1008-1008/1009-...-2010/2011-2011/2012
Tìm x:
x . (1/2+1/3+1/4+. . .+1/2011+1/2012)
2012/1+2011/2+2010/3+2009/4+ . . . +2/2011+1/2012
=1
Thực hiện phép tính (1/2+1/3+1/4+1/5+1/6+...+1/2012)/(2011/1+2010/2+2009/3+...+1/2011)
Tim x:
x . (1/2+1/3+1/4+. . .+1/2011+1/2012)
2012/1 +2011/2+2010/3+2009/4+2/2011+1/2012
=1
(1/2+1/3+1/4+...+1/2013).x=2012/1+2011/2+...+2/2011+1/2012
mình đang cần gấp.Ngày 26 tháng 2 năm 2018 là mình phải nộp rồi
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(1-1/2)*(1-1/3)*(1-1/4)*(1-1/5)*.........*(1-1/2011)*(1-2012)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2012}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{2011}{2012}\)
\(=\frac{1\cdot2\cdot...\cdot2011}{2\cdot3\cdot...\cdot2012}\)
\(=\frac{1}{2012}\)
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\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2011}\right)\left(1-\frac{1}{2012}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2010}{2011}.\frac{2011}{2012}\)
\(=\frac{1.2.3...2010.2011}{2.3.4...2011.2012}\)
\(=\frac{1}{2012}\)
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Ta có \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).\left(1-\frac{1}{5}\right)......\left(1-\frac{1}{2012}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.....\frac{2010}{2011}.\frac{2011}{2012}\)
\(=\frac{1.2.3.4.5...2010.2011}{2.3.4.5.6....2011.2012}\)
\(=\frac{1}{2012}\)
Vậy.....
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Xem thêm câu trả lời
Câu 1:So sánh M= 1/1.2+1/2.3+...+1/49.50 với 1
Câu 2: Tính. B=1+2+2^2+2^3+...+2^2008/1-2^2009
Câu 3.Tính. B=1/2+1/6+1/12+1/20+1/30+...+1/9900
Câu 4.Tính. 1/1.3+1/3.5+1/5.7+...+1/2009.2011
Câu 5. So sánh:
A=2011+2012/2012+2013
Và B=2011/2012+2011/2012+2012/2013
Câu 6: Tìm x biết :.(x/7+0,25)=-1/28
Tính giá trị của biểu thức:
\(\dfrac{\dfrac{1}{2013}+\dfrac{2}{2012}+\dfrac{3}{2011}+...+\dfrac{2011}{3}+\dfrac{2012}{2}+\dfrac{2013}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}}\)
\(A=\dfrac{\dfrac{1}{2013}+\dfrac{2}{2012}+\dfrac{3}{2011}+...+\dfrac{2011}{3}+\dfrac{2012}{2}+\dfrac{2013}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2014}}\)
\(A=\dfrac{1+\left(\dfrac{1}{2013}+1\right)+\left(\dfrac{2}{2012}+1\right)+\left(\dfrac{3}{2011}+1\right)+...+\left(\dfrac{2012}{2}+1\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2014}}\)
\(A=\dfrac{\dfrac{2014}{2014}+\dfrac{204}{2013}+\dfrac{2014}{2012}+\dfrac{2014}{2011}+...+\dfrac{2014}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2014}}\)
\(A=\dfrac{2014\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2014}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2014}}=2014\)
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mình ko chắc đúng nha !
Số số hạng của tử là :
(2013-1):1+1=2013(số hạng)
\(\dfrac{\dfrac{1}{2013}+\dfrac{2}{2012}+\dfrac{3}{2011}+.....+\dfrac{2011}{3}+\dfrac{2012}{2}+\dfrac{2013}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2014}}\)
\(=\dfrac{\dfrac{1}{2013}+1+\dfrac{2}{2012}+1+....+\dfrac{2012}{2}+1+\dfrac{2013}{1}-2012}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2014}}\)
\(=\dfrac{\dfrac{2014}{2013}+\dfrac{2014}{2012}+....+\dfrac{2014}{2}+1}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2014}}\)
\(=2014\left(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2014}}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2014}}\right)\)
=2014
Mình ghi thêm ở cái dâu bằng thứ 2 cuối cùng trên tử có ghi trừ 2012 là do tử có 2013 hạng tử mà mình chỉ cộng 1 cho 2012 hạng tử nên phải trừ đi 2012
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P=(1-1/2).(1-1/3).(1-1/4).(1-1/5)....(1-1/2011).(1-1/2012)
\(P=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2011}\right)\left(1-\frac{1}{1012}\right)\)
\(P=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2011}{2012}\)
\(P=\frac{1\cdot2\cdot3\cdot...\cdot2011}{2\cdot3\cdot4\cdot...\cdot2012}\)
\(P=\frac{1}{2012}\)
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