\(\frac{1}{5.8}+\frac{1}{8.11}+......+\frac{1}{302.305}\)
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=1/3.3(1/2.5-1/5.8-1/8.11-...-1/302.305)
=1/3.(3/2.5-3/5.8-3/8.11-...-3/302.305)
=1/3(1/2-1/5-1/5-1/8-1/8-1/11-...-1/302-1/305)
=1/3[(1/2-1/305)+(1/5-1/5)+...+(1/302-1/302)
=1/3*(1/2-1/305)=1/3*(305/610-1/610)=1/3*304/610=152/915
hình như mình làm sai hoặc sai đề , sao số lớn ghê
\(\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right]\)
\(\frac{1}{3}\left[\frac{1}{2}-\frac{1}{20}\right]=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}\)
mk đầu tiên đó
3S=3/2.5+3/5.8+3/8.11+...+3/101.104
3S=1/2-1/5+1/5-1/8+1/8-1/11+...+1/101-1/104
3S=1/2-1/104
S=51/104:3
S=17/104
Vậy S=17/104
\(S=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+........+\frac{1}{101.104}\)
\(\Rightarrow3S=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+.......+\frac{1}{101.104}\right)\)
\(=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+........+\frac{3}{101.104}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.........+\frac{1}{101}-\frac{1}{104}\)
\(=\frac{1}{2}-\frac{1}{104}\)
\(=\frac{51}{104}\)
\(\Rightarrow S=\frac{51}{104}:3=\frac{51}{104}.\frac{1}{3}\)
\(=\frac{7}{104}\)
VẬY \(S=\frac{7}{104}\)
A = \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)
A = \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\)
A = \(\frac{1}{2}-\frac{1}{98}\)
A = \(\frac{24}{49}\)
Vậy A = \(\frac{24}{49}\)
~~~
#Sunrise
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)
\(=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}+\frac{3}{95.98}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{98}\right)\)
\(=\frac{1}{3}.\frac{24}{49}=\frac{8}{49}\)
\(1-\frac{1}{2\cdot5}-\frac{1}{5\cdot8}-\frac{1}{8\cdot11}-...-\frac{1}{92\cdot95}\)
\(=1-\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{92\cdot95}\right)\)
\(=1-\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{2}{92\cdot95}\right)\)
\(=1-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}\right)\)
\(=1-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{95}\right)\)
\(=1-\frac{1}{3}\cdot\frac{93}{190}\)
\(=1-\frac{31}{190}\)
\(=\frac{159}{190}\)
\(1-\frac{1}{2.5}-\frac{1}{5.8}-\frac{1}{8.11}-...-\frac{1}{92.95}\)
\(=1-\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}\right)\)
\(=1-\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}\right)\)
\(=1-\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}\right)\)
\(=1-\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{95}\right)\)
\(=1-\frac{1}{3}.\frac{93}{190}\)
\(=1-\frac{31}{190}\)
\(=\frac{159}{190}\)
Đặt C = \(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{2015.2018}\)
\(\Rightarrow3C=\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{2015.2018}\)
\(\Rightarrow3C=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{2015}-\frac{1}{2018}\)
\(\Rightarrow3C=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)
\(\Rightarrow C=\frac{504}{1009}:3=\frac{168}{1009}\)
Vậy \(C=\frac{168}{1009}\)
\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{x+3}=-\frac{3}{10}\)
\(\Leftrightarrow1\cdot10=-3\left(x+3\right)\)
\(\Leftrightarrow10=-3x-9\)
\(\Leftrightarrow10+9=-3x\)
\(\Leftrightarrow19=-3x\)
\(\Leftrightarrow x=-\frac{19}{3}\)
Đề sai à -.-
\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{6}\)
=> \(\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{1}{6}\)
=> \(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{1}{6}:\frac{1}{3}\)
=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{1}{6}\cdot3=\frac{1}{2}\)
=> \(\frac{1}{x+3}=\frac{1}{5}-\frac{1}{2}=-\frac{3}{10}\)
=> \(10=-3\left(x+3\right)\)
=> 10 = -9x - 9
=> 10 + 9x + 9 = 0
=> 19 + 9x = 0
=> 9x = -19
=> x = -19/9
\(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{10300}=\frac{1}{x}\)
=> \(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{100\cdot103}=\frac{1}{x}\)
=> \(\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{100\cdot103}\right)=\frac{1}{x}\)
=> \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{1}{x}\)
=> \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{103}\right)=\frac{1}{x}\)
=> \(\frac{101}{618}=\frac{1}{x}\)
=> \(101x=618\)
=> \(x=\frac{618}{101}\)
Vậy : ...
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{10300}=\frac{1}{x}\)
\(\Rightarrow\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{100.103}=3.\frac{1}{x}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{103}=3.\frac{1}{x}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{103}=3.\frac{1}{x}\)
\(\Rightarrow\frac{1}{x}.3=\frac{101}{206}\)
\(\Rightarrow\frac{1}{x}=\frac{101}{618}\)
\(\Rightarrow x=\frac{618}{101}\)
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{302.305}\)
=\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{302}-\frac{1}{305}\)
=\(\frac{1}{5}-\frac{1}{305}\)
=\(\frac{12}{61}\)
tick cho mik nha
Nhầm bạn ơi
Lúc đầu nhân 3 xong lúc cuối chia cho 3 nha
Đáp án là \(\frac{12}{61}:3=\frac{4}{61}\) sr bạn