1/3 + 1/6 +1/10 + ... +1/x.(x+1):2=1998/2000
Những câu hỏi liên quan
1/3+ 1/6 +1/10 +..................+2/x*(x+1) =1998/2000
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
BT3: Tìm x, biết
20) \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(x+1\right):2}=\dfrac{1998}{2000}\)
\(\dfrac{1.2}{3.2}+\dfrac{1.2}{6.2}+.....+\dfrac{1.2}{2\left(x+1\right):2.2}\)=\(\dfrac{1998}{2000}\)
\(\dfrac{2}{6}+\dfrac{2}{12}+.....+\dfrac{2}{x\left(x+1\right)}\)=\(\dfrac{1998}{2000}\)
\(2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{1998}{2000}\)
\(2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{1998}{2000}\)
\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1998}{2000}:2\)
\(\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{999}{2000}\)
\(\dfrac{1}{x+1}=\dfrac{1}{2000}\)
suy ra x+1=2000
suy ra x=2000-1=1999
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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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Tính các tổng sau:
1,S1=1+(-3)+5+(-7)+...+1997+(-1999)
2,S2=1+(-2)+(-3)+4+5+(-6)+(-7)+8+...+1997+(-1998)+(-1999)+2000
3,S3= 2-4+6-8+...+1998-2000
4,S4=2-4-6+8+10-12-14+16+...+1994-1996-1998+2000+2009
Các bạn ơi giúp mình với ạ,mình đang cần gấp !!!!
1, S1 = (-2) + (-2) +..+ (-2).
Có SS (-2) là :
(1997 - 1) : 4 +1 = 500 (số ).
Tổng số (-2) là: 500 x (-2) = (-1000)
Bạn chờ mình làm tiếp nha
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Các bạn ơi làm giúp mình vs ạ,mình đang cần gấp lắm rồi!!!!HELP MEEEEEEEEEEEEEE
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a) dfrac{x+1}{35}+dfrac{x+3}{33}dfrac{x+5}{31}+dfrac{x+7}{29}Hd: cộng thêm 1 vào các hạng tửb) dfrac{x-10}{1994}+dfrac{x-8}{1996}+dfrac{x-6}{1998}+dfrac{x-4}{2000}+dfrac{x-2}{2002}dfrac{x-2002}{2}+dfrac{x-2000}{4}+dfrac{x-1998}{6}+dfrac{x-1996}{8}+dfrac{x-1994}{10}Hd: trừ đi 1 vào các hạng tửc) dfrac{x-1991}{9}+dfrac{x-1993}{7}+dfrac{x-1995}{5}+dfrac{x-1997}{3}+dfrac{x-1999}{1}dfrac{x-9}{1991}+dfrac{x-7}{1993}+dfrac{x-5}{1995}+dfrac{x-3}{1997}+dfrac{x-1}{1999}Hd: trừ đi 1 vào các hạng tửd) dfrac...
Đọc tiếp
a) \(\dfrac{x+1}{35}\)+\(\dfrac{x+3}{33}\)=\(\dfrac{x+5}{31}\)+\(\dfrac{x+7}{29}\)Hd: cộng thêm 1 vào các hạng tửb) \(\dfrac{x-10}{1994}\)+\(\dfrac{x-8}{1996}\)+\(\dfrac{x-6}{1998}\)+\(\dfrac{x-4}{2000}\)+\(\dfrac{x-2}{2002}\)=\(\dfrac{x-2002}{2}\)+\(\dfrac{x-2000}{4}\)+\(\dfrac{x-1998}{6}\)+\(\dfrac{x-1996}{8}\)+\(\dfrac{x-1994}{10}\)Hd: trừ đi 1 vào các hạng tử
c) \(\dfrac{x-1991}{9}\)+\(\dfrac{x-1993}{7}\)+\(\dfrac{x-1995}{5}\)+\(\dfrac{x-1997}{3}\)+\(\dfrac{x-1999}{1}\)=\(\dfrac{x-9}{1991}\)+\(\dfrac{x-7}{1993}\)+\(\dfrac{x-5}{1995}\)+\(\dfrac{x-3}{1997}\)+\(\dfrac{x-1}{1999}\)Hd: trừ đi 1 vào các hạng tửd) \(\dfrac{x-8}{15}\)+\(\dfrac{x-74}{13}\)+\(\dfrac{x-67}{11}\)+\(\dfrac{x-64}{9}\)=10Chú ý: 10=1+2+3+4e) \(\dfrac{x-1}{13}\)-\(\dfrac{2x-13}{15}\)=\(\dfrac{3x-15}{27}\)-\(\dfrac{4x-27}{29}\)Hd: thêm hoặc bớt 1 vào các hạng tử
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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x+1/2009 + x+2/2008 + x+3/2007 + x=10/2000 + x==11/1999 + x+12/1998
\(\dfrac{x+1}{2009}+\dfrac{x+2}{2008}+\dfrac{x+3}{2007}=\dfrac{x+10}{2000}+\dfrac{x+11}{1999}+\dfrac{x+12}{1998}\)
\(\Rightarrow\left(\dfrac{x+1}{2009}+1\right)+\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)=\left(\dfrac{x+10}{2000}+1\right)+\left(\dfrac{x+11}{1999}+1\right)+\left(\dfrac{x+12}{1998}+1\right)\)
\(\Rightarrow\dfrac{x+2010}{2009}+\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}=\dfrac{x+2010}{2000}+\dfrac{x+2010}{1999}+\dfrac{x+2010}{1998}\)\(\Rightarrow\dfrac{x+2010}{2009}+\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}-\dfrac{x+2010}{2000}-\dfrac{x+2010}{1999}-\dfrac{x+2010}{1998}=0\)\(\Rightarrow\left(x+2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2007}-\dfrac{1}{2000}-\dfrac{1}{1999}-\dfrac{1}{1998}\right)=0\)\(\Rightarrow x+2010=0\Rightarrow x=-2010\)
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