Theo hệ quả của định lí Thalès ta có:

\(\frac{C A}{C E} = \frac{C B}{C D} = \frac{A B}{D E} = \frac{5}{15} = \frac{1}{3}\)

Hay:

⚡\(\frac{C B}{C D} = \frac{1}{3}\) suy ra \(\frac{x}{7 , 2} = \frac{1}{3}\).

Vậy \(x = \frac{7 , 2. \&\text{nbsp}; 1}{3} = 2 , 4\)

⚡\(\frac{C A}{C E} = \frac{1}{3}\) suy ra \(\frac{3}{y} = \frac{1}{3}\)

Vậy \(y = \frac{3.3}{1} = 9\).

\(\frac{2 x - 50}{50} + \frac{2 x - 51}{49} + \frac{2 x - 52}{48} + \frac{2 x - 53}{47} + \frac{2 x - 100}{25} + \frac{- 100}{25} = 0\)

\(\frac{2 x - 50}{50} + \frac{2 x - 51}{49} + \frac{2 x - 52}{48} + \frac{2 x - 53}{47} + \frac{2 x - 100}{25} + \left(\right. - 4 \left.\right) = 0\)

\(\frac{2 x - 50}{50} - 1 + \frac{2 x - 51}{49} - 1 + \frac{2 x - 52}{48} - 1 + \frac{2 x - 53}{47} - 1 + \frac{2 x - 100}{25} = 0\)

\(\frac{2 x - 100}{50} + \frac{2 x - 100}{49} + \frac{2 x - 100}{48} + \frac{2 x - 100}{47} + \frac{2 x - 100}{25} = 0\)

\(\left(\right. 2 x - 100 \left.\right) . \left(\right. \frac{1}{50} + \frac{1}{49} + \frac{1}{48} + \frac{1}{47} + \frac{1}{25} \left.\right) = 0\)

\(2 x - 100 = 0\) (Do \(\frac{1}{50} + \frac{1}{49} + \frac{1}{48} + \frac{1}{47} + \frac{1}{25} \neq 0\))

\(x = 50\).

Vậy phương trình đã cho có nghiệm \(x = 50\).

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