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a)=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}\)
\(=1-\frac{1}{2009}\)
\(=\frac{2008}{2009}\)
b) =\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{94}-\frac{1}{97}\)
\(=1-\frac{1}{97}\)
=\(\frac{96}{97}\)
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2008.2009}\) \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2008}-\frac{1}{2009}\)
= 1 - 1/2009
= 2008/2009
b) 3/1.4 + 3/4.7 + 3/7.10 + .... + 3/94.97
= 1- 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + .... + 1/94 - 1/97
= 1 - 1/97
= 96/97
a; \(\dfrac{-1}{n}\) - \(\dfrac{1}{n+a}\)
= \(\dfrac{-n-a-n}{n.\left(n+a\right)}\)
= \(\dfrac{-2n-a}{n.\left(n+a\right)}\)
b; \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ....+ \(\dfrac{1}{2007.2008}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2007}-\dfrac{1}{2008}\)
= \(\dfrac{1}{1}\) - \(\dfrac{1}{2008}\)
= \(\dfrac{2007}{2008}\)
c; \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)
= \(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)
= \(\dfrac{1}{1}\) - \(\dfrac{1}{97}\)
= \(\dfrac{96}{97}\)
Câu 2:
a: =>-11/12x=-1/6-3/4=-2/12-9/12=-11/12
=>x=1
b: =>x-42=57-x-50=7-x
=>2x=49
hay x=49/2
d: =>x+1=3 hoặc x+1=-3
=>x=2 hoặc x=-4
e: =>2x+3=5 hoặc 2x+3=-5
=>2x=2 hoặc 2x=-8
=>x=1 hoặc x=-4
Đề bài :
a) dãy các phân số trên có phải theo quy luật ko ?
b) tính tổng các phân số của dãy trên
1) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=1-\dfrac{1}{50}\)
\(=\dfrac{49}{50}\)
2) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{37.39}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{37}-\dfrac{1}{39}\)
\(=\dfrac{1}{3}-\dfrac{1}{39}\)
\(=\dfrac{13}{39}-\dfrac{1}{39}=\dfrac{12}{39}=\dfrac{4}{13}\)
3) \(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{73.76}\)
\(=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{73}-\dfrac{1}{76}\)
\(=\dfrac{1}{4}-\dfrac{1}{76}\)
\(=\dfrac{19}{76}-\dfrac{1}{76}=\dfrac{18}{76}=\dfrac{9}{38}\)
1)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ =1-\dfrac{1}{50}\\ =\dfrac{49}{50}\)
2)
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{37.39}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\\ =\dfrac{1}{3}-\dfrac{1}{39}\\ =\dfrac{13}{39}-\dfrac{1}{39}\\ =\dfrac{12}{39}=\dfrac{4}{13}\)
3) \(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{73.76}\\ =\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{73}-\dfrac{1}{79}\\ =\dfrac{1}{4}-\dfrac{1}{79}\\ =\dfrac{75}{316}\)
\(\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...............+\dfrac{2}{2008.2009}\)
\(=2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+................+\dfrac{1}{2008.2009}\right)\)
\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.................+\dfrac{1}{2008}-\dfrac{1}{2009}\right)\)
\(=2\left(1-\dfrac{1}{2009}\right)\)
\(=2.\dfrac{2008}{2009}=\dfrac{4016}{2009}\)
a)\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{46}-\frac{1}{56}\)
=\(1-\frac{1}{56}=\frac{55}{56}\)
b)\(A.\frac{1}{3}=\frac{1}{3}.\left(\frac{3}{1.2}+\frac{3}{2.3}+....+\frac{3}{99.100}\right)\)
= \(\frac{1}{3}A=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{3}{99.100}\)
=> \(\frac{1}{3}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
=> \(\frac{1}{3}A=1-\frac{1}{100}=\frac{99}{100}\)
=> \(A=\frac{99}{100}.3=\frac{297}{100}\)
c)\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)
=\(1-\frac{1}{103}=\frac{102}{103}\)
d) \(\frac{3}{5}C=\frac{3}{5}.\left(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\right)\)
=\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\)
=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{103}\)
=\(1-\frac{1}{103}=\frac{102}{103}\)
=>\(C=\frac{102}{103}.\frac{5}{3}=\frac{170}{103}\)
e) \(\frac{4}{7}D=\frac{4}{7}.\left(\frac{7}{1.5}+\frac{7}{5.9}+...+\frac{7}{101.105}\right)\)
=\(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{101.105}\)
=\(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\)
=\(1-\frac{1}{105}=\frac{104}{105}\)
=< D=\(\frac{104}{105}.\frac{7}{4}=\frac{26}{15}\)
a) \(P=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+...\dfrac{10}{46.56}\)
\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...\dfrac{1}{46}-\dfrac{1}{56}\)
\(P=1-\dfrac{1}{56}\)
\(P=\dfrac{55}{56}\)
b) \(A=\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+...+\dfrac{3}{99.100}\)
\(A=3\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)
\(A=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=3\left(1-\dfrac{1}{100}\right)\)
\(A=3.\dfrac{99}{100}\)
\(A=\dfrac{297}{100}\)
c) \(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)
\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(B=1-\dfrac{1}{103}\)
\(B=\dfrac{102}{103}\)
d) \(C=\dfrac{5}{1.4}+\dfrac{5}{4.7}+\dfrac{5}{7.10}+...+\dfrac{5}{100.103}\)
\(C=\dfrac{5}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\right)\)
\(C=\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(C=\dfrac{5}{3}\left(1-\dfrac{1}{103}\right)\)
\(C=\dfrac{5}{3}.\dfrac{102}{103}\)
\(C=\dfrac{170}{103}\)
e) \(D=\dfrac{7}{1.5}+\dfrac{7}{5.9}+\dfrac{7}{9.13}+...+\dfrac{7}{101.105}\)
\(D=\dfrac{7}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{101.105}\right)\)
\(D=\dfrac{7}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{101}-\dfrac{1}{105}\right)\)
\(D=\dfrac{7}{4}\left(1-\dfrac{1}{105}\right)\)
\(D=\dfrac{7}{4}.\dfrac{104}{105}\)
\(D=\dfrac{26}{15}\)
a, \(\frac{9}{1.2}+\frac{9}{2.3}+...+\frac{9}{99.100}\)
=9.(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\))
= 9(1 -\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\))
=9(1-\(\frac{1}{100}\))
A=\(\frac{891}{100}\)
b, \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{27.30}\)
=1-(\(\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{27}-\frac{1}{30}\))
=1-\(\frac{1}{30}\)
B=\(\frac{29}{30}\)
a) \(\dfrac{9}{1.2}+\dfrac{9}{2.3}+...+\dfrac{9}{99.100}\)
\(=9\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)
\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=9\left(1-\dfrac{1}{100}\right)\)
\(=9.\dfrac{99}{100}\)
\(=\dfrac{891}{100}\)
b) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{27.30}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{27}-\dfrac{1}{30}\)
\(=1-\dfrac{1}{30}\)
\(=\dfrac{29}{30}\)
Giải:
b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2008}-\dfrac{1}{2009}\)
\(=\dfrac{1}{1}-\dfrac{1}{2009}\)
\(=\dfrac{2008}{2009}\)
c) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{4}{7.10}+...+\dfrac{3}{94.97}\)
\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)
\(=\dfrac{1}{1}-\dfrac{1}{97}\)
\(=\dfrac{96}{97}\)
Vậy ...
Các câu sau tương tự
b, \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2008.1009}\)
\(=\)\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2008}-\dfrac{1}{2009}\)
\(=\dfrac{1}{1}-\dfrac{1}{2009}=\dfrac{2009}{2009}-\dfrac{1}{2009}=\dfrac{2008}{2009}\)