Tìm x,biết x/2x5 + x/5x8 + x/8x11 + x/11x14 +...+ x/32x35 = 33/70
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\(\Leftrightarrow x\cdot\dfrac{1}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{32\cdot35}\right)=\dfrac{33}{70}\)
=>\(x\cdot\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{32}-\dfrac{1}{35}\right)=\dfrac{33}{70}\)
=>\(x\cdot\dfrac{1}{3}\cdot\dfrac{33}{70}=\dfrac{33}{70}\)
=>x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x}{2\times5}+\dfrac{x}{5\times8}+\dfrac{x}{8\times11}+\dfrac{x}{11\times14}+...+\dfrac{x}{32\times35}=\dfrac{33}{70}\)
\(\dfrac{x}{3}\cdot\left(\dfrac{3}{2\times5}+\dfrac{3}{5\times8}+\dfrac{3}{8\times11}+\dfrac{3}{11\times14}+...+\dfrac{3}{32\times35}\right)=\dfrac{33}{70}\)
\(\dfrac{x}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{32}-\dfrac{1}{35}\right)=\dfrac{33}{70}\)
\(\dfrac{x}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{35}\right)=\dfrac{33}{70}\)
\(\dfrac{x}{3}\cdot\dfrac{33}{70}=\dfrac{33}{70}\)
\(\dfrac{x}{3}=\dfrac{33}{70}:\dfrac{33}{70}\)
\(\dfrac{x}{3}=1\)
\(x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{3x}{2\cdot5}+\frac{3x}{5\cdot8}+\frac{3x}{8\cdot11}+\frac{3x}{11\cdot14}=\frac{1}{21}\)
\(\Leftrightarrow x\cdot\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right)=\frac{1}{21}\)
\(\Leftrightarrow x\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)
\(\Leftrightarrow x\cdot\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)
\(\Leftrightarrow x\cdot\frac{3}{7}=\frac{1}{21}\)
\(\Leftrightarrow x=\frac{1}{21}:\frac{3}{7}=\frac{1}{21}\cdot\frac{7}{3}=\frac{7}{63}=\frac{1}{9}\)
Vậy: \(x=\frac{1}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\times\left(\frac{1}{5\times8}+\frac{1}{8\times11}+....+\frac{1}{97\times100}+x\right)=\frac{319}{100}\)
\(\Rightarrow\left(\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+...+\frac{3}{97\times100}\right)+3\times x=\frac{319}{100}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{97}-\frac{1}{100}+3\times x=\frac{319}{100}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{100}+3\times x=\frac{319}{100}\)
\(\Rightarrow\frac{19}{100}+3\times x=\frac{319}{100}\)
\(\Rightarrow3\times x=\frac{319}{100}-\frac{19}{100}\)
\(\Rightarrow3\times x=3\)
\(\Rightarrow x=3:3\)
\(\Rightarrow x=1\)
Vậy x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{27}{480}\)
\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{27}{480}.\frac{1}{3}\)
\(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{3}{160}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{1}{160}\)
\(\Rightarrow\frac{1}{x+3}=\frac{31}{160}\)
\(\Rightarrow160=31x+93\)
\(\Rightarrow31x=67\)
\(\Rightarrow x=\frac{67}{31}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : A=3/2x5+3/5x8+3/8x11+3/11x14+3/14x17+3/17x20
=> A=1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17+1/17-1/20
=> A=1/2-1/20
=> A=9/20
Vậy A=9/20
đó
![](https://rs.olm.vn/images/avt/0.png?1311)
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![](https://rs.olm.vn/images/avt/0.png?1311)
= 5-2/2x5+8-5/5x8+11-8/8x11+14-11/11x14
=(1/2-1/5)+(1/5-1/8)+(1/8-1/11)+(1/11-1/14)
=(1/2+1/5+1/8+1/11)-(1/5+1/8+1/11+1/14)
=1/2-1/14
=3/7
Vậy B=3/7
\(\dfrac{x}{2\cdot5}+\dfrac{x}{5\cdot8}+\dfrac{x}{8\cdot11}+\dfrac{x}{11\cdot14}+...+\dfrac{x}{32\cdot35}=\dfrac{33}{70}\)
\(\Leftrightarrow x\left(\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\dfrac{1}{11\cdot14}+...+\dfrac{1}{32\cdot35}\right)=\dfrac{33}{70}\)
\(\Rightarrow x\cdot\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+...+\dfrac{3}{32\cdot35}\right)=\dfrac{33}{70}\)
\(\Leftrightarrow x\cdot\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{32}-\dfrac{1}{35}\right)=\dfrac{33}{70}\)
\(\Leftrightarrow x\cdot\dfrac{1}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{35}\right)=\dfrac{33}{70}\)
\(x\cdot\dfrac{1}{3}\cdot\dfrac{33}{70}=\dfrac{33}{70}\)
\(x=\dfrac{33}{70}:\dfrac{33}{70}:\dfrac{1}{3}\)
\(x=3\)
x = 3