Question

Tìm số nguyên x, biết:

\(\dfrac{x}{6}+\dfrac{x}{10}+\dfrac{x}{15}+\dfrac{x}{21}+\dfrac{x}{28}+\dfrac{x}{36}+\dfrac{x}{45}+\dfrac{x}{55}+\dfrac{x}{66}+\dfrac{x}{78}=\dfrac{200}{39}\)

Answer 1

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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