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a, \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\) \(=-88\)
\(x+\dfrac{206}{100}=\dfrac{-5}{176}\)
\(x=\dfrac{-5}{176}-\dfrac{206}{100}\)
\(x=\dfrac{-9198}{4400}\)
a) \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=90-89\)
\(\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)
\(\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=\dfrac{1}{2}\)
\(x+\dfrac{206}{100}=5\)
\(x=5-\dfrac{206}{100}\)
\(x=\dfrac{147}{50}\)
Vậy \(x=\dfrac{147}{50}\)
11: \(=\left(1+\dfrac{1}{98}-1-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{98}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\right)=0\)
12: \(=\dfrac{7}{17}+\dfrac{10}{17}\cdot\left(\dfrac{-6+5}{10}\right)^2\)
\(=\dfrac{7}{17}+\dfrac{10}{17}\cdot\dfrac{1}{100}=\dfrac{7}{17}+\dfrac{1}{170}=\dfrac{71}{170}\)
|2x - 1|.\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{1996.1997}\right)\)= 1996
|2x - 1|.\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1996}-\dfrac{1}{1997}\right)\)= 1996
|2x - 1|.\(\left(1-\dfrac{1}{1997}\right)\)= 1996
|2x - 1|. \(\dfrac{1996}{1997}\)= 1996
|2x - 1| = 1996 : \(\dfrac{1996}{1997}\)
|2x - 1| = 1996 . \(\dfrac{1997}{1996}\)
|2x - 1| = 1997
2x - 1 = ± 1997
TH1:
2x -1 = 1997
2x = 1997 +1
2x= 1998
x= 1998:2
x=999
TH2:
2x-1= -1997
2x= -1997+1
2x= -1996
x= -1996:2
x= -998
Vậy x ∈ {999; -998}
\(\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)..............\left(1-\dfrac{2}{99.100}\right)\)
\(=\left(\dfrac{6}{2.3}-\dfrac{2}{2.3}\right).\left(\dfrac{12}{3.4}-\dfrac{2}{3.4}\right)..............\left(\dfrac{9900}{99.100}-\dfrac{2}{99.100}\right)\)
\(=\dfrac{4}{2.3}.\dfrac{10}{3.4}..........................\dfrac{9898}{99.100}\)
\(=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}..............\dfrac{98.101}{99.100}\)
\(=\dfrac{1.2.3.....98}{2.3......99}.\dfrac{4.5.6.....101}{3.4.......100}\)
\(=\dfrac{1}{99}.\dfrac{101}{3}=\dfrac{101}{297}\)
a) Để phân số \(\dfrac{3}{n-2}\) là số nguyên thì n - 2 \(⋮\) 3
\(\Rightarrow\) n - 2 \(\in\) Ư(3)
\(\Rightarrow\) n - 2 \(\in\){3; -3; 1;-1}
n \(\in\){5; -1; 3; 2}
c) \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+......+\dfrac{1}{28.29}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{29}-\dfrac{1}{30}\)
\(=\dfrac{1}{3}-\dfrac{1}{30}\)
\(=\dfrac{10}{30}-\dfrac{1}{30}\)
\(=\dfrac{9}{30}\)
=\(\dfrac{3}{10}\)
P\(=\dfrac{3}{\left(1.2\right)^2}+\dfrac{5}{\left(2.3\right)^2}+.....+\dfrac{4033}{\left(2016.2017\right)^2}\) \(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+.......+\dfrac{4033}{2016^2.2017^2}\) \(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+....+\dfrac{1}{2016^2}-\dfrac{1}{2017^2}\) =1\(-\dfrac{1}{2017^2}\) Do `1\(-\dfrac{1}{2017^2}\) <1\(\Rightarrow\) P<1 ( ĐPCM)
P = \(\dfrac{3}{\left(1.2\right)^2}+\dfrac{5}{\left(2.3\right)^2}+\dfrac{7}{\left(3.4\right)^2}+...+\dfrac{4033}{\left(2016.2017\right)^2}\)
P = \(\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{4033}{\left(2016.2017\right)^2}\)
P = \(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{2016^2}-\dfrac{1}{2017^2}\)
P = \(1-\dfrac{1}{2017^2}\)
⇒ P < 1
⇒ ĐPCM
\(a,\dfrac{3}{4}-1\dfrac{1}{2}+0,5:\dfrac{5}{12}.\)
\(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}.\)
\(=\dfrac{3}{4}-\dfrac{6}{4}+\dfrac{1}{2}.\dfrac{12}{5}.\)
\(=-\dfrac{3}{4}+\dfrac{12}{10}.\)
\(=-\dfrac{3}{4}+\dfrac{6}{5}.\)
\(=-\dfrac{15}{20}+\dfrac{24}{20}=\dfrac{9}{20}.\)
Vậy.....
\(b,\left(-2\right)^2-1\dfrac{5}{27}.\left(-\dfrac{3}{2}\right)^3.\)
\(=4-1\dfrac{5}{27}.\left(-\dfrac{27}{8}\right).\)
\(=4-\dfrac{32}{27}.\left(-\dfrac{27}{8}\right).\)
\(=4-\left(-4\right).\)
\(=4+4=8.\)
Vậy.....
\(c,\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}.\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}.\)
\(=\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{99}-\dfrac{1}{99}\right)-\dfrac{1}{100}.\)
\(=\dfrac{1}{2}+0+0+...+0-\dfrac{1}{100}.\)
\(=\dfrac{1}{2}-\dfrac{1}{100}.\)
\(=\dfrac{50}{100}-\dfrac{1}{100}=\dfrac{49}{100}.\)
Vậy.....
a) \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}+\dfrac{1}{195}\)
\(=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-.....+\dfrac{1}{13}-\dfrac{1}{15}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{15}\right)\)
\(=\dfrac{1}{2}.\dfrac{4}{15}=\dfrac{2}{15}\)
b) \(\dfrac{4}{9}:\left(-\dfrac{1}{7}\right)+6\dfrac{5}{9}:\left(-\dfrac{1}{7}\right)\)
\(=\dfrac{4}{9}.\left(-7\right)+\dfrac{59}{9}\left(-7\right)\)
\(=-7\left(\dfrac{4}{9}+\dfrac{59}{9}\right)\)
\(=-7.7=-49\)
c) \(\left(3\dfrac{2}{5}-2\dfrac{2}{5}\right).\left(-\dfrac{5}{3}\right)+3.\left(2\dfrac{1}{2}:\dfrac{1}{2}\right)\)
\(=\left(\dfrac{17}{5}-\dfrac{12}{5}\right).\left(-\dfrac{5}{3}\right)+3.5\)
\(=-\dfrac{5}{3}+15=13\dfrac{1}{3}\)
d) \(1\dfrac{13}{5}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}+1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
\(=\dfrac{2}{7}+78\dfrac{8}{15}:\dfrac{47}{24}\)
( bạn tự tính nốt câu này nha ! )
\(\Rightarrow\left(1+1+...+1\right)+2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{n\left(n+1\right)}\right)\)[có (n-1) số 1]
\(\Rightarrow\left(n-1\right)+2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)
\(\Rightarrow\left(n-1\right)+2\left(\dfrac{1}{2}-\dfrac{1}{n+1}\right)\)
\(\Rightarrow\left(n-1\right)+\left(1-\dfrac{2}{n+1}\right)\)
\(\Rightarrow n-\dfrac{2}{n+1}\)
\(\Rightarrow\dfrac{n\left(n+1\right)}{n+1}-\dfrac{2}{n+1}\)
\(\Rightarrow\dfrac{n^2+n-2}{n+1}\)