a: \(3\left(1-2x\right)^2+6x\left(\dfrac{4}{3}x+1\right)=3x\)
=>\(3\left(4x^2-4x+1\right)+8x^2+6x-3x=0\)
=>\(12x^2-12x+3+8x^2+3x=0\)
=>\(20x^2-9x+3=0\)
\(\text{Δ}=\left(-9\right)^2-4\cdot20\cdot3=81-80\cdot3=81-240=-159< 0\)
=>Phương trình vô nghiệm
b: -3x(x-6)-x+6=0
=>\(-3x\left(x-6\right)-\left(x-6\right)=0\)
=>\(\left(x-6\right)\left(-3x-1\right)=0\)
=>(x-6)(3x+1)=0
=>\(\left[{}\begin{matrix}x-6=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-\dfrac{1}{3}\end{matrix}\right.\)
c: \(4\left(\dfrac{1}{2}x-3\right)^2-9\left(\dfrac{5}{3}x+2\right)^2=0\)
=>\(\left[2\left(\dfrac{1}{2}x-3\right)\right]^2-\left[3\left(\dfrac{5}{3}x+2\right)^2\right]=0\)
=>\(\left(x-6\right)^2-\left(5x+6\right)^2=0\)
=>\(\left(x-6-5x-6\right)\left(x-6+5x+6\right)=0\)
=>\(6x\left(-4x-12\right)=0\)
=>\(6\cdot\left(-4\right)\cdot x\left(x+3\right)=0\)
=>x(x+3)=0
=>\(\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
