x.(\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+\(\dfrac{1}{15}\)+\(\dfrac{1}{21}\)+\(\dfrac{1}{28}\)+\(\dfrac{1}{36}\)+\(\dfrac{1}{45}\)+\(\dfrac{1}{55}\)+\(\dfrac{1}{66}\)+\(\dfrac{1}{78}\))=\(\dfrac{220}{39}\)

x.\(\dfrac{20}{39}\)=\(\dfrac{220}{39}\)

x=\(\dfrac{220}{39}\):\(\dfrac{20}{39}\)

x=11

m

lppkokoko

\(\frac{x}{6}+\frac{x}{10}+\frac{x}{15}+\frac{x}{21}+\frac{x}{28}+\frac{x}{36}+\frac{x}{45}+\frac{x}{55}+\frac{x}{66}+\frac{x}{78}+\frac{x}{78}=\frac{220}{39}\)

\(\Leftrightarrow x\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}+\frac{1}{66}+\frac{1}{78}+\frac{1}{78}\right)=\frac{220}{39}\)

\(\Leftrightarrow x\cdot\frac{20}{39}=\frac{220}{39}\Rightarrow x=11\)

\(\frac{x}{6}+\frac{x}{10}+\frac{x}{15}+\frac{x}{21}+\frac{x}{28}+\frac{x}{36}+\frac{x}{45}+\frac{x}{55}+\frac{x}{66}+\frac{x}{78}+\frac{x}{78}=\frac{220}{39}\)

\(=>x=\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}+\frac{1}{66}+\frac{1}{78}+\frac{1}{78}=\frac{220}{39}\)

\(x\cdot\frac{20}{39}=\frac{220}{39}\)

\(x=\frac{220}{39}:\frac{20}{39}=11\)

\(x.\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{78}\right)=\dfrac{220}{39}\)

\(x.\dfrac{20}{39}=\dfrac{220}{39}\)

\(x=\dfrac{220}{39}:\dfrac{20}{39}\)

x\(=11\)

⇔2x(13.4+14.5+..........+112.13)=22039

      \(\dfrac{x}{6}+\dfrac{x}{10}+\dfrac{x}{15}+\dfrac{x}{21}+\dfrac{x}{28}+\dfrac{x}{36}+\dfrac{x}{45}+\dfrac{x}{55}+\dfrac{x}{66}+\dfrac{x}{78}=\dfrac{220}{39}\)

⇔ \(x.\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}+\dfrac{1}{55}+\dfrac{1}{66}+\dfrac{1}{78}\right)\) \(=\) \(\dfrac{220}{39}\)

⇔ \(x.\dfrac{20}{39}=\dfrac{220}{39}\)

⇔ \(x=11\)

\(\dfrac{x}{6}+\dfrac{x}{10}+\dfrac{x}{15}+........+\dfrac{x}{78}=\dfrac{220}{39}\)

\(\Leftrightarrow\dfrac{2x}{12}+\dfrac{2x}{20}+........+\dfrac{2x}{156}=\dfrac{220}{39}\)

\(\Leftrightarrow2x\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+..........+\dfrac{1}{12.13}\right)=\dfrac{220}{39}\)

\(\Leftrightarrow2x\left(\dfrac{1}{3}-\dfrac{1}{13}\right)=\dfrac{220}{39}\)

\(\Leftrightarrow2x.\dfrac{10}{39}=\dfrac{220}{39}\)

\(\Leftrightarrow x.\dfrac{20}{39}=\dfrac{220}{39}\)

\(\Leftrightarrow x=11\)

Vậy ...

mk sửa 2020 thành 220

???

x.(1/6+1/10+1/15+1/21+1/28+1/36+1/45+1/55+1/66+1/78)=220/39

x.20/39=220/39

x=220/39:20/39

x=11

Gọi biểu thức là A

\(A=\dfrac{2x}{12}+\dfrac{2x}{20}+\dfrac{2x}{30}+....+\dfrac{2x}{156}=\dfrac{200}{39}\)

Ta có công thức :

\(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)

Áp dụng công thức trên, ta có :

\(A=\dfrac{2x}{3.4}+\dfrac{2x}{4.5}+\dfrac{2x}{5.6}+....+\dfrac{2x}{12.13}\)

\(A=2x.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{12}-\dfrac{1}{13}\right)\)

\(A=2x.\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)

\(A=2x.\left(\dfrac{10}{39}\right)=\dfrac{200}{39}\)

\(A=2x=\dfrac{200}{39}:\dfrac{10}{39}\)

\(2x=20\)

\(\Rightarrow x=10\)

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