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Ta có :\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)
= \(\dfrac{1}{1.2.3}.\dfrac{2}{2}+\dfrac{1}{2.3.4}.\dfrac{2}{2}+...+\dfrac{1}{98.99.100}.\dfrac{2}{2}\)
= \(\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{98.99.100}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}+0+0+...+0+\dfrac{-1}{99.100}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{2}+\dfrac{-1}{9900}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{4850}{9900}+\dfrac{-1}{9900}\right)\)
= \(\dfrac{1}{2}.\dfrac{4849}{9900}\)
= \(\dfrac{4849}{19800}\)
\(D=\dfrac{1}{2}+\dfrac{-1}{5}+\dfrac{-5}{7}+\dfrac{1}{6}+\dfrac{-3}{35}+\dfrac{1}{3}+\dfrac{1}{41}\)
\(D=\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{3}\right)+\left(\dfrac{-1}{5}+\dfrac{-5}{7}+\dfrac{-3}{35}\right)+\dfrac{1}{41}\)
\(D=1+-1+\dfrac{1}{41}\)
\(D=0+\dfrac{1}{41}\)
\(D=\dfrac{1}{41}\)
\(C=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)+\left(\dfrac{-3}{4}+\dfrac{-1}{36}+\dfrac{-2}{9}\right)+\dfrac{1}{57}\)
\(=\dfrac{5+9+1}{15}+\dfrac{-27-1-8}{36}+\dfrac{1}{57}\)
=1/57
\(E=\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{4}{35}+\dfrac{2}{7}\right)+\dfrac{1}{127}=\dfrac{1}{127}\)
\(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{899}{900}\)
\(A=\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot...\cdot\dfrac{29\cdot31}{30\cdot30}\)
\(A=\dfrac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot31}{\left(2\cdot3\cdot4\cdot...\cdot30\right)^2}\)
\(A=\dfrac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot31}{\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot30}\)
\(A=\dfrac{1\cdot31}{30}=\dfrac{31}{30}\)
Ta có : \(\dfrac{1}{101}>\dfrac{1}{300}\)
...
\(\dfrac{1}{299}>\dfrac{1}{300}\)
Do đó :
\(\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{300}>\dfrac{1}{300}+\dfrac{1}{300}..+\dfrac{1}{300}\)
\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{300}>\dfrac{200}{300}=\dfrac{2}{3}\)
Vậy...
\(A=\dfrac{1}{1.2}-\dfrac{1}{1.2.3}+\dfrac{1}{2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{3.4}-\dfrac{1}{3.4.5}+\dfrac{1}{99.100}-\dfrac{1}{99.100.101}\)
\(A=\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)-\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{99.100.101}\right)\)
\(A=\left(1-\dfrac{1}{100}\right)-\left(\dfrac{\dfrac{1}{1.2}-\dfrac{1}{100.101}}{2}\right)\)
Bấm máy nha
\(B=\dfrac{5}{1.2.3.4}+\dfrac{5}{2.3.4.5}+\dfrac{5}{3.4.5.6}+...+\dfrac{5}{98.99.100.101}\)
\(B=\dfrac{5}{3}.\left(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{4-1}{1.2.3.4}+\dfrac{5-2}{2.3.4.5}+...+\dfrac{101-98}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{4}{1.2.3.4}-\dfrac{1}{1.2.3.4}+\dfrac{5}{2.3.4.5}-\dfrac{2}{2.3.4.5}+...+\dfrac{101}{98.99.100.101}-\dfrac{98}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{1}{1.2.3}-\dfrac{1}{99.100.101}\right)\)
\(B=\dfrac{5}{3}.\dfrac{166649}{999900}\approx0,3\)
a) Ta có:
3A= \(1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\left(1\right)\)
A= \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\left(2\right)\)
Lấy (1) - (2) ta được:
1-\(\dfrac{1}{3^{100}}\)
b) Ta xét:
\(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{2}{1.2.3},...,\dfrac{1}{37.38}-\dfrac{1}{38.39}=\dfrac{2}{37.38.39}\)
Ta có:
2B=\(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+..+\dfrac{2}{37.38.39}\)
=\(\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)+\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+..+\left(\dfrac{1}{37.38}-\dfrac{1}{38.39}\right)\)
=\(\dfrac{1}{1.2}-\dfrac{1}{38.39}=\dfrac{740}{38.39}=\dfrac{370}{741}\)
Vậy \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+..+\dfrac{2}{37.38.39}\)
=\(\dfrac{370}{741}\)
Nếu bn cảm thấy mk đúng tick cho mk nhé!
a)\(\dfrac{3}{10}\)-x=\(\dfrac{25}{30}\)-\(\dfrac{4}{30}\)
\(\dfrac{3}{10}-x=\dfrac{7}{10}\)
x = \(\dfrac{3}{10}-\dfrac{7}{10}\)
x=\(\dfrac{-4}{10}\)
b)\(\dfrac{-5}{8}+x=\dfrac{4}{9}-\dfrac{63}{9}\)
\(\dfrac{-5}{9}+x=\dfrac{-59}{9}\)
\(x=\dfrac{-59}{9}-\dfrac{-5}{9}\)
\(x=\dfrac{-64}{9}\)
c)=>2.18=(x-3).(x-3)
=>36=(x-3)\(^2\)
=>6\(^2\)=(x-3)\(^2\)
6= x-3
x=6+3=9
a, A= 1/2. (2/1.2.3+2/2.3.4+2/3.4.5+...+2/18.19.20) A=1/2. (1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+...+1/18.19-1/19.20) A=1/2. (1/1.2-1/19.20) A=1/2. 189/380 A= 189/760
Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.
\(linh_1=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}\)
\(linh_1=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}\right)\)
\(linh_1=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{4.5}\right)\)
\(linh_1=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=\dfrac{1}{2}.\dfrac{9}{20}=\dfrac{9}{40}\)
\(linh_2=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{8.9.10}\)
\(linh_2=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\right)\)\(linh_2=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{9.10}\right)\)
\(linh_2=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{90}\right)=\dfrac{1}{2}.\dfrac{22}{45}=\dfrac{11}{45}\)
a/ \(G=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}\)
\(\Leftrightarrow2G=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}\)
\(\Leftrightarrow2G=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}\)
\(\Leftrightarrow2G=\dfrac{1}{1.2}-\dfrac{1}{4.5}\)
\(\Leftrightarrow2G=\dfrac{1}{2}-\dfrac{1}{20}\)
\(\Leftrightarrow2G=\dfrac{9}{20}\)
\(\Leftrightarrow G=\dfrac{9}{40}\)
b/ \(H=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+.....+\dfrac{1}{8.9.10}\)
\(\Leftrightarrow2H=\dfrac{2}{1.2.3}+\dfrac{2}{3.4.5}+.....+\dfrac{2}{8.9.10}\)
\(\Leftrightarrow2H=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+.....+\dfrac{1}{8.9}-\dfrac{1}{9.10}\)
\(\Leftrightarrow2H=\dfrac{1}{1.2}-\dfrac{1}{9.10}\)
\(\Leftrightarrow2H=\dfrac{1}{2}-\dfrac{1}{90}\)
\(\Leftrightarrow2H=\dfrac{22}{45}\)
\(\Leftrightarrow H=\dfrac{22}{90}\)
\(A=\left(\dfrac{-1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{4}{35}+\dfrac{2}{7}\right)+\dfrac{1}{127}\)
\(A=\left(\dfrac{-9-2-7}{18}\right)+\left(\dfrac{21+4+10}{35}\right)+\dfrac{1}{127}\)
\(A=-1+1+\dfrac{1}{127}\)
\(A=\dfrac{1}{127}\)
\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\)
\(\dfrac{1}{4}B=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.4}+\dfrac{1}{3.4.5.4}+...+\dfrac{1}{98.99.100.4}\)
\(\dfrac{1}{4}B=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.\left(5-1\right)}+\dfrac{1}{3.4.5.\left(6-2\right)}+...+\dfrac{1}{98.99.100.\left(101-97\right)}\)
\(\dfrac{1}{4}B=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5-1.2.3.4}+\dfrac{1}{3.4.5.6-2.3.4.5}+...+\dfrac{1}{98.99.100.101-97.98.99.100}\)
\(\dfrac{1}{4}B=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}-\dfrac{1}{1.2.3.4}+\dfrac{1}{3.4.5.6}-\dfrac{1}{2.3.4.5}+...+\dfrac{1}{98.99.100.101}-\dfrac{1}{97.98.99.100}\)
\(\dfrac{1}{4}B=\dfrac{1}{98.99.100.101}\)
\(B=\dfrac{1}{98.99.100.101}.4=\dfrac{1}{98.99.25.101}\)
tick cho mk nha
bài tự làm 100%
co gì chưa đc thì coi lại nha