Question

Tìm x biết: \(x+3+\dfrac{4-3a^2}{a^2-9}=\dfrac{5}{2a^2+6a}\) ( a \(\ne\) 0, a \(\ne\pm\)3 )

Answer 1

ta có : \(x+3+\dfrac{4-3a^2}{a^2-9}=\dfrac{5}{2a^2+6a}\)

\(\Leftrightarrow x+3=\dfrac{5}{2a^2+6a}-\dfrac{4-3a^2}{a^2-9}\)

\(\Leftrightarrow x+3=\dfrac{5}{2a\left(a+3\right)}-\dfrac{4-3a^2}{\left(a+3\right)\left(a-3\right)}\) \(\Leftrightarrow x+3=\dfrac{5\left(a-3\right)-2a\left(4-3a^2\right)}{2a\left(a+3\right)\left(a-3\right)}\) \(\Leftrightarrow x+3=\dfrac{5a-15-8a+6a^3}{2a\left(a+3\right)\left(a-3\right)}=\dfrac{6a^3-3a-15}{2a\left(a+3\right)\left(a-3\right)}\)

\(\Leftrightarrow x=\dfrac{6a^3-3a-15}{2a\left(a+3\right)\left(a-3\right)}-3=\dfrac{6a^3-3a-15-3.2a\left(a^2-9\right)}{2a\left(a+3\right)\left(a-3\right)}\)

\(\Leftrightarrow x=\dfrac{6a^3-3a-15-6a^3+54a}{2a\left(a+3\right)\left(a-3\right)}=\dfrac{51a-15}{2a\left(a^2-9\right)}\)