$\dfrac{5}{4\cdot3}+\dfrac{3}{4\cdot11}+\dfrac{3}{7\cdot11}+\dfrac{9}{7\cdot23}=\dfrac{2x}{69}$
$\dfrac{5}{12}+\dfrac{3}{44}+\dfrac{3}{77}+\dfrac{9}{161}=\dfrac{2x}{69}$
$\dfrac{5}{12}+\dfrac{3}{28}+\dfrac{9}{161}=\dfrac{2x}{69}$
$\dfrac{11}{21}+\dfrac{9}{161}=\dfrac{2x}{69}$
$\dfrac{280}{483}=\dfrac{2x}{69}$
$\dfrac{40}{69}=\dfrac{2x}{69}$
$40=2x$
$x=20$
Đây không phải toán lớp 2, em nhé.
\(\frac{5}{4.3}\) + \(\frac{3}{4.11}\) + \(\frac{3}{7.11}\) + \(\frac{9}{7.23}\) = \(\frac{2x}{69}\)
\(\frac{5}{2.4.3}\) + \(\frac{3}{2.4.11}\) + \(\frac{3}{2.7.11}\) + \(\frac{9}{2.7.23}\) = \(\frac{2x}{2.69}\)
\(\frac{5}{8.3}\) + \(\frac{3}{8.11}\) + \(\frac{3}{14.11}\) + \(\frac{3}{14.23}\) = \(\frac{x}{69}\)
\(\frac13-\frac18\) + \(\frac18-\frac{1}{11}\) + \(\frac{1}{11}\) - \(\frac{1}{14}\) + \(\frac{1}{14}\) - \(\frac{1}{23}\) = \(\frac{x}{69}\)
\(\frac13\) - \(\frac{1}{23}\) = \(\frac{x}{69}\)
\(\frac{20}{69}\) = \(\frac{x}{69}\)
20 = \(x\)
Vậy \(x\) = 20
