Giải pt: $\frac{1}{x}$ + $\frac{1}{x+\frac{3}{2} }$ = $\frac{5}{9}$

Đáp án:

\(\left[ \begin{array}{l}
x = 3\\
x =  - \dfrac{9}{{10}}
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
DK:x \ne \left\{ { - \dfrac{3}{2};0} \right\}\\
\dfrac{1}{x} + \dfrac{1}{{x + \dfrac{3}{2}}} = \dfrac{5}{9}\\
 \to \dfrac{1}{x} + 1:\left( {\dfrac{{2x + 3}}{2}} \right) = \dfrac{5}{9}\\
 \to \dfrac{1}{x} + \dfrac{2}{{2x + 3}} = \dfrac{5}{9}\\
 \to \dfrac{{2x + 3 + 2x}}{{x\left( {2x + 3} \right)}} = \dfrac{5}{9}\\
 \to 9\left( {4x + 3} \right) = 5x\left( {2x + 3} \right)\\
 \to 36x + 27 = 10{x^2} + 15x\\
 \to 10{x^2} - 21x - 27 = 0\\
 \to 10{x^2} - 30x + 9x - 27 = 0\\
 \to 10x\left( {x - 3} \right) + 9\left( {x - 3} \right) = 0\\
 \to \left( {x - 3} \right)\left( {10x + 9} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 3\\
x =  - \dfrac{9}{{10}}
\end{array} \right.
\end{array}\)