a: \(x^2+2x=4x^2-1\)
=>\(4x^2-1-x^2-2x=0\)
=>\(3x^2-2x-1=0\)
=>\(3x^2-3x+x-1=0\)
=>(x-1)(3x+1)=0
=>\(\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
b:
\(4\left(x-1\right)^2=9\left(x+3\right)^2\)
=>\(\left(3x+9\right)^2=\left(2x-2\right)^2\)
=>\(\left(3x+9\right)^2-\left(2x-2\right)^2=0\)
=>(3x+9+2x-2)(3x+9-2x+2)=0
=>(5x+7)(x+11)=0
=>\(\left[{}\begin{matrix}5x+7=0\\x+11=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{7}{5}\\x=-11\end{matrix}\right.\)
c: \(2x^2-5x+3=0\)
=>\(2x^2-2x-3x+3=0\)
=>2x(x-1)-3(x-1)=0
=>(x-1)(2x-3)=0
=>\(\left[{}\begin{matrix}x-1=0\\2x-3=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
d: x(x-1)(x+1)(x+2)-3=0
=>\(x\left(x+1\right)\left(x-1\right)\left(x+2\right)-3=0\)
=>\(\left(x^2+x\right)\left(x^2+x-2\right)-3=0\)
=>\(\left(x^2+x\right)^2-2\left(x^2+x\right)-3=0\)
=>\(\left(x^2+x-3\right)\left(x^2+x+1\right)=0\)
mà \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}>0\forall x\)
nên \(x^2+x-3=0\)
=>\(x=\dfrac{-1\pm\sqrt{13}}{2}\)
