\(\Leftrightarrow x^2+\dfrac{5}{3}x-3=0\Leftrightarrow x^2+\dfrac{5}{3}x+\dfrac{25}{36}-\dfrac{133}{36}=0\Leftrightarrow\left(x-\dfrac{5}{6}\right)^2=\dfrac{133}{36}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{133}+5}{6}\\x=\dfrac{-\sqrt{133}+5}{6}\end{matrix}\right.\)
Nghiệm ra xấu, em xem lại đề như nào nha!
\(\Leftrightarrow\left(20x-4\right)^2-\left(9x+15\right)^2=0\)
\(\Leftrightarrow\left(20x-4-9x-15\right)\left(20x-4+9x+15\right)=0\)
=>(11x-19)(29x+11)=0
=>x=19/11 hoặc x=-11/29
<=> (20x-4)^2 - (15+9x)^2 =0
<=> (11x-19)(29x+11) = 0
<=> x = 19/11;x=-11/29
X = - \(\dfrac{11}{29}\)