1/5.8+1/8.11+1/11.14|+...+1/605.608
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\(\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)
\(\dfrac{3}{2}A=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{97-94}{94.97}\)
\(\dfrac{3}{2}A=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{97}{94.97}-\dfrac{94}{94.97}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{97}=\dfrac{96}{97}\)
⇒ A = \(\dfrac{96}{97}:\dfrac{3}{2}=\dfrac{64}{97}\)
Câu B cách làm tương tự, thắc mắc gì bạn cứ hỏi nhé.
Đặt 1/5.8 + 1/8.11 +...+ 1 /x (x+3) = A
3A = 3/5.8 + 3/8.11 +...+ 3/x (x+3)
3A = 1/5 - 1/8 + 1/8 - 1/11 +...+ 1/x - 1/x+3
3A = 1/5 - 1/x + 3
3A = ( 3+x)-5/5x +15
A =[ ( 3+ x ) - 5 / 5x + 15 ] : 3
A = x + ( - 2 ) / 5x + 15
Ta có :
A + 27/480
= x + ( - 2 ) / 5x + 15
=> x + ( - 2 ) = 27
=> 5x + 15 = 480
* Làm nốt *
#Louis
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{27}{480}\)
\(=\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{27}{480}\)
\(=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{27}{480}\)
\(=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{27}{480}\)
\(=\frac{1}{5}-\frac{1}{x+3}=\frac{27}{480}.3\)
\(=\frac{1}{5}-\frac{1}{x+3}=\frac{81}{480}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{81}{480}=\frac{15}{480}=\frac{1}{32}\)
\(\Rightarrow x+3=32\)
\(\Rightarrow x=32-3=29\)
Ta có: \(\dfrac{k}{x.\left(x+k\right)}=\dfrac{x+k-x}{x.\left(x+k\right)}=\dfrac{1}{x}-\dfrac{1}{x+k}\)
nên áp dụng ta có:
\(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x.\left(x+3\right)}\)
\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\)
\(=\dfrac{1}{5}-\dfrac{1}{x+3}\)
Nên $\dfrac{1}{3}.\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x.\left(x+3\right)}\right)=\dfrac{1}{3}.(\dfrac{1}{5}-\dfrac{1}{x+3})$
Đến đây là làm được rồi nha
S = 1/2.5 +1/5.8 +1/8.11+1/11.14+1/14.17+1/17.20
S=1/3.(1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17+1/17-1/20)
S=1/3.(1/2-1/20)
S=1/3.(10/20-1/20)
S=1/3.9/20
S= 3/20
k nha
\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+1\right)}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+1}\right)=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{303}{1540}\Rightarrow\frac{1}{x+1}=\frac{1}{308}\)
=> x + 1 = 380 => x = 308 - 1 => x = 307
Vậy x = 307
=1/3(3/5.8+3/8.11+............+1/x(x+3)=101/1540
=.1/3(1/5.8+1/8.11+......1/x(x+3)=101/1540
=1/3(1/5-1/8+1/8-1/11+...........1/x-1/x+3=101/1540
=>1/3(1/5-1/x+3)=101/1540
=>1/5-1/x+3=101/1540 chia 1/3 =303/1540
=>1/x+3= 1/308
...........
Gọi A=1/5.8+1/8.11+1/11.14+...+1/605.608
A×3=3/5.8+3/8.11+3/11.14+...+3/605.608
A×3=1/5-1/8+1/8-1/11+1/11-1/14+...+1/605-1/608
A×3=1/5-1/608
A×3=603/3040
A=603/3040÷3
A=201/3040