1/3+1/6+1/10+...+2/x(x+1)=1999/2000. ; b)2+2/3+2/6+2/12+...+2/x(x+1)=3980/1991
Những câu hỏi liên quan
tìm x, biết :
a) 1/( x+1 ).( x+2 ) + 1/( x+2 ).( x+3 ) + 1/( x+3 ).( x+4 ) + ... + 1/( x+1999).( x+2000) = 1/x+200 = 1/5
b) 2/( x+1 ).( x+3 ) + 3/( x+3 ) .( x+6 ) + 4/( x+6 ).( x+10 ) + ... + 10/( x+45 ).( x+55 ) + 1/ x+55 = 1/6
b1 :tính
a) N (1/2 - 1). (1/3 - 1). (1/4 - 1) .... (1/2016 - 1). (1/2017 - 1)
b) P (-1/1/2). (-1/1/3). (-1/1/4) ... (-1/1/2015). (2016/01/01)
c) Q 1/13 + 3 / 13,23 + 3 / 23,33 + ... + 3 / 2013,2023
d) R 1 / 2.017,2016-1 / 2.016,2015-1 / 2015,2014 - ... -. 1 / (x + 1999) (x + 2000)
b2 : tìm x, biết :
a) 1 / (x + 1). (x + 2) + 1 / (x + 2). (x + 3) + 1 / (x + 3). (x + 4) + ... + 1 / (x + 1999). (x + 2000) 1 / x + 200 1/5
b) 2 / (x + 1). (x + 3) + 3 / (x + 3). (x + 6) + 4 / (x +...
Đọc tiếp
b1 :tính
a) N = (1/2 - 1). (1/3 - 1). (1/4 - 1) .... (1/2016 - 1). (1/2017 - 1)
b) P = (-1/1/2). (-1/1/3). (-1/1/4) ... (-1/1/2015). (2016/01/01)
c) Q = 1/13 + 3 / 13,23 + 3 / 23,33 + ... + 3 / 2013,2023
d) R = 1 / 2.017,2016-1 / 2.016,2015-1 / 2015,2014 - ... -. 1 / (x + 1999) (x + 2000)
b2 : tìm x, biết :
a) 1 / (x + 1). (x + 2) + 1 / (x + 2). (x + 3) + 1 / (x + 3). (x + 4) + ... + 1 / (x + 1999). (x + 2000) = 1 / x + 200 = 1/5
b) 2 / (x + 1). (x + 3) + 3 / (x + 3). (x + 6) + 4 / (x + 6). (x + 10) + ... + 10 / (x + 45). (x + 55) + 1 / x + 55 = 1/6
( chú ý : / là phần nha )
a,x-10/1994+x-8/1996+x-6/1998+x-4/2000+x-2/2002=x-2002/2+x-2000/4+x-1998/6+x-1996/8+x-1994/10
b,x-1991/9+x-1993/7+x-1995/5+x-1997/3+x-1999/1=x-9/1991+x-7/1993+x-5/1995+x-3/1997+x-1/1999
c,x-1/13-2x-13/15=3x-15/27-4x-27/29
Tính S = 1 x 2 - 2 x 3 + 3 x 4 - 4 x 5 + 5 x 6 - 6 x 7 + ... - 1998 x 1999 + 1999 x 2000.
Tìm x biết:
1) \(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}\)
2) \(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
1) \(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}\)
<=> \(\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)
<=> \(x+1=0\) (do 1/2 + 1/3 + 1/4 - 1/5 - 1/6 khác 0)
<=> \(x=-1\)
Vậy...
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\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
<=> \(\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1\)
<=> \(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)
<=> \(\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)
<=> \(x+2010=0\) (do 1/2009 + 1/2008 + 1/2007 - 1/2000 - 1/1999 - 1/1998 khác 0)
<=> \(x=-2010\)
Vậy....
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Tìm x biết:
1) \(\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}=\dfrac{x+1}{5}+\dfrac{x+1}{6}\)
2) \(\dfrac{x+1}{2009}+\dfrac{x+2}{2008}+\dfrac{x+3}{2007}=\dfrac{x+10}{2000}+\dfrac{x+11}{1999}+\dfrac{x+12}{1998}\)
1,
x+1/2+x+1/3+x+1/4-x+1/5-x+1/6=0
(x+1)(1/2+1/3+1/4-1/5-1/6)=0
vì 1/2+1/3+1/4-1/5-1/6 khác 0
suy ra x+1=0 suy ra x=-1
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19 tháng 12 2021 lúc 21:19
Bài 12: Tính :
a) A = 1 + (-3) + 5 + ( - 7) +….+ 17 + ( -19);
b) B = (- 2) + 4 + (-6) + 8 + …+ ( - 18) + 20;
c) C = 1 + (-2) + 3 + (-4) + ….+ 1999 + ( - 2000) + 2001;
Bài 13: Tìm số nguyên x, biết:
a) –x + 20 = -(-15) –(+8) + 13
b) –(-10) + x = -13 + (-9) + (-6)
Bài 13:
a: =>20-x=15-8+13=20
hay x=0
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Mong các bạn giúp mình bài này
Khi khai triển và ước lượng số hạng đồng dạng của
P(x)=(1−x+x^2−x^3+...−x^1999+x^2000)(1+x+x^2+x^3+...+x^1999+x^2000)(1−x+x^2−x^3+...−x^1999+x^2000)(1+x+x^2+x^3+...+x^1999+x^2000) ta có thể viết P(x) dưới dạng
P(x)= a0+a1.x+a2.x^2+a3.x^3+...+a4000.x^4000
Tính a2001
Đặt \(A=1-x+x^2-x^3+...-x^{1999}+x^{2000}\)
\(B=1+x+x^2+x^3+...+x^{1999}+x^{2000}\)
Ta có : \(\left(x^2-1\right).P\left(x\right)=\left(x+1\right)A\left(x-1\right)B\)
\(=\left(x^{2001}+1\right)\left(x^{2001}-1\right)\)
\(=\left(x^{2001}\right)^2-1=\left(x^2\right)^{2001}-1^{2001}\)
\(=\left(x^2-1\right)\left(x^{4000}+x^{3998}+x^{3996}+...+x^2+1\right)\)
\(\Rightarrow P\left(x\right)=x^{4000}+x^{3998}+...+x^2+1\)
Theo đề bài ta có : \(P\left(x\right)=a_o+a_1x+...+a_{4000}x^{4000}\)
Do đó : hệ số chẵn sẽ = 1, hệ số lẻ = 0
\(\Rightarrow a_{2001}=0\)
Chúc bạn học tốt !!
x+1/2009+x+2/2008+x+3/2007=x+10/2000+x+11/1999+x+12/1998
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
\(\Rightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+1010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\right)=\left(x+2010\right)\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}\right)\)
\(\Rightarrow x+2010=0\) vì \(0< \frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}< \frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}\)
\(\Rightarrow x=-2010\)
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Bài giải
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
\(\Rightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+10}{2000}+1\right)+\left(\frac{x+11}{1999}+1\right)+\left(\frac{x+12}{1998}+1\right)\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-(\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998})=0\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
\(\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)
Vì \(\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)\ne0\) nên \(x+2010=0\)
\(x=0-2010=-2010\)
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Phan Uyên Nhi
Bạn bấm vào câu hỏi tương tự rồi tham khảo nha !
Có rất nhiều bài giống bài của bạn hỏi đó !
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