\(\dfrac{1}{9}-\dfrac{1}{1009}\)
Những câu hỏi liên quan
\(A=\left(\dfrac{1}{7}+\dfrac{1}{23}-\dfrac{1}{1009}\right):\left(\dfrac{1}{23}+\dfrac{1}{7}-\dfrac{1}{1009}+\dfrac{1}{7}.\dfrac{1}{23}.\dfrac{1}{1009}\right)+1:\left(30.1009-160\right)\)
tìm giá trị biểu thức sau :
A=(\(\dfrac{1}{7}\)+\(\dfrac{1}{23}-\dfrac{1}{1009}\)):(\(\dfrac{1}{23}+\dfrac{1}{7}-\dfrac{1}{1009}+\dfrac{1}{7}.\dfrac{1}{23}.\dfrac{1}{1009}\))+1:(30.1009-160)
\(A=\dfrac{2016^2+1^2}{2016\cdot1}+\dfrac{2015^2+2^2}{2015\cdot1}+\dfrac{2014^2+3^2}{2014\cdot3}+...+\dfrac{1009^2+1008^2}{1009\cdot1008}\)
và \(B=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}\)Tìm A/B
\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{2n\left(2n+2\right)}=\dfrac{1009}{4038}\) Tìm n?
Ta có \(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{2n\left(2n+2\right)}=\dfrac{1009}{4038}\)
\(\Leftrightarrow\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2n\left(2n+2\right)}=\dfrac{1009}{2019}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2n}-\dfrac{1}{2n+2}=\dfrac{1009}{2019}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2n+2}=\dfrac{1009}{2019}\)
\(\Leftrightarrow\dfrac{n}{2n+2}=\dfrac{1009}{2019}\)
\(\Leftrightarrow2019n=1009\left(2n+2\right)\)
\(\Leftrightarrow2019n=2018n+2018\)
\(\Leftrightarrow n=2018\)
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\(\dfrac{x+1}{2014}+\dfrac{x+2}{2013}=...+\dfrac{x+1007}{1008}=\dfrac{x+1008}{1007}+\dfrac{x+1009}{1006}+...+\dfrac{x+2014}{1}\)
\(\dfrac{x+1}{2014}+\dfrac{x+2}{2013}+.....+\dfrac{x+1007}{1008}=\dfrac{x+1008}{1007}+\dfrac{x+1009}{1006}+........+\dfrac{x+2014}{1}\)\(\Leftrightarrow\left(\dfrac{x+1}{2014}+1\right)+\left(\dfrac{x+2}{2013}+1\right)+...+\left(\dfrac{x+1007}{1008}+1\right)=\left(\dfrac{x+1008}{1007}+1\right)+\left(\dfrac{x+1009}{1006}+1\right)+...+\left(\dfrac{x+2014}{1}+1\right)\)\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}+...+\dfrac{x+1007}{1008}=\dfrac{x+2015}{1007}+\dfrac{x+1009}{1006}+...+\dfrac{x+2014}{1}\)\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}+...+\dfrac{x+2015}{1008}-\dfrac{x+1008}{1007}-\dfrac{x+2015}{1006}-...-\dfrac{x+2015}{1}=0\)\(\Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1008}-\dfrac{1}{1007}-\dfrac{1}{1006}-...-1\right)=0\)\(\Leftrightarrow x+2015=0\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1008}-\dfrac{1}{1007}-\dfrac{1}{1006}-...-1>0\right)\)\(\Leftrightarrow x=-2015\)
Vậy x=-2015
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so sánh \(\dfrac{1}{\sqrt{1.2017}}+\dfrac{1}{\sqrt{2.2016}}+...+\dfrac{1}{\sqrt{2017.1}}\) với \(\dfrac{2017}{1009}\)
Viết các đa thức sau thành tích1. ( 3x + 2)2 - 42. 4x2 - 25y23. 4x2- 494. 8z3 + 275. dfrac{9}{25}x4 - dfrac{1}{4}6. x32 - 17. 4x2 + 4x + 18. x2 - 20x + 1009. y4 -14y2 + 4910. 125x3 - 64y3
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Viết các đa thức sau thành tích
1. ( 3x + 2)2 - 4
2. 4x2 - 25y2
3. 4x2- 49
4. 8z3 + 27
5. \(\dfrac{9}{25}\)x4 - \(\dfrac{1}{4}\)
6. x32 - 1
7. 4x2 + 4x + 1
8. x2 - 20x + 100
9. y4 -14y2 + 49
10. 125x3 - 64y3
1. ( 3x + 2)2 - 4
= (3x+2-2)(3x+2+2)
= 3x(3x+4)
2. 4x2 - 25y2
= (2x-5y)(2x+5y)
3. 4x2- 49
=(2x-7)(2x+7)
4. 8z3 + 27
=(2z+3)(4x2-6z+9)
5. \(\dfrac{9}{25}x^4-\dfrac{1}{4}\)
= \((\dfrac{3}{5}x^2-\dfrac{1}{2})(\dfrac{3}{5}x^2+\dfrac{1}{2})\)
6. x32 - 1
=(x16-1)(x16+1)
7. 4x2 + 4x + 1
=(2x+1)2
8. x2 - 20x + 100
=(x-10)2
9. y4 -14y2 + 49
=(y2-7)2
10. 125x3 - 64y3
= (5x-4y)(25x2+20xy+16y2)
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1) \(\left(3x+2\right)^2-4=\left(3x+2+2\right)\left(3x+2-2\right)=3x\left(3x+4\right)\)
2) \(4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)
3) \(4x^2-49=\left(2x-7\right)\left(2x+7\right)\)
4) \(8z^3+27=\left(2z+3\right)\left(4z^2-6z+9\right)\)
5) \(\dfrac{9}{25}x^4-\dfrac{1}{4}=\left(\dfrac{3}{5}x^2-\dfrac{1}{2}\right)\left(\dfrac{3}{5}x^2+\dfrac{1}{2}\right)\)
6) \(x^{32}-1=\left(x^{16}-1\right)\left(x^{16}+1\right)\)
\(=\left(x^8-1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
7) \(4x^2+4x+1=\left(2x+1\right)^2\)
8) \(x^2-20x+100=\left(x-10\right)^2\)
9) \(y^4-14y^2+49=\left(y^2-7\right)^2\)
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2 tháng 4 2017 lúc 22:35
Cho :S=\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+.......+\dfrac{1}{2013}-\dfrac{1}{2014}+\dfrac{1}{2015}\) và P=\(\dfrac{1}{1008}+\dfrac{1}{1009}+......+\dfrac{1}{2014}+\dfrac{1}{2015}\) Tính \(\left(S-P\right)^{2016}\)
Ta có:
\(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2013}-\dfrac{1}{2014}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2014}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-2\left(1+\dfrac{1}{2}+...+\dfrac{1}{2014}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{2017}\right)\)
\(=\dfrac{1}{1008}+\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}\)
Mà \(P=\dfrac{1}{1008}+\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}\)
\(\Rightarrow S=P\Rightarrow S-P=0\)
\(\Rightarrow\left(S-P\right)^{2016}=0^{2016}=0\)
Vậy \(\left(S-P\right)^{2016}=0\)
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\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+.........+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}\)
\(B=\dfrac{1}{1007}+\dfrac{1}{1008}+\dfrac{1}{1009}+.........+\dfrac{1}{2013}\)
tính \(\left(A-B\right)^{2013}\)
B=\(\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\)
=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1006}+\dfrac{1}{1007}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)\)- \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1006}\right)\)
=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1006}+\dfrac{1}{1007}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)\)-2\(\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\)
=1-\(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...-\dfrac{1}{2012}+\dfrac{1}{2013}\)=S
( A-B)2013 =0
Chúc ban học tốt nhé...!
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