c: =>(x-1)(x+1)=0

hay \(x\in\left\{1;-1\right\}\)

plss

a,

\(=\dfrac{3}{4x}.\left(x-3\right)\left(x+3\right)\)=0

\(\left\{{}\begin{matrix}\dfrac{3}{4x}=0\\x-3=0\\x+3=0\end{matrix}\right.\)

=>\(x=\left\{3,-3\right\}\)

b,

\(x^3-16x=0\\x\left(x^2-16\right)\\ x\left(x-4\right)\left(x+4\right)\)

\(\left\{{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)

=>\(x=\left\{-4,0,4\right\}\)

d,

\(3x^3-27x=0\\ 3x\left(x^2-9\right)=0\\ 3x\left(x-3\right)\left(x+3\right)=0\)

\(\left\{{}\begin{matrix}3x=0\\x-3=0\\x+3=0\end{matrix}\right.\)

=>\(x=\left\{-3,0,3\right\}\)

e,

\(x^2+\left(x+1\right)+2x\left(x+1\right)=0\\ x\left(x+1\right)\left(x+2\right)=0\)

\(\left\{{}\begin{matrix}x=0\\x+1=0\\x+2=0\end{matrix}\right.\)

=>\(x=\left\{-2,-1,0\right\}\)

f,

\(x\left(2x-3\right)-2\left(3-2x\right)=0\\ \left(2x-3\right)\left(x+2\right)=0\)

\(\left\{{}\begin{matrix}2x-3=0\\x+2=0\end{matrix}\right.\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)

khó quá       

\(x=0\)

\(3x^3-75x=0\Leftrightarrow3x\left(x^2-25\right)=0\Leftrightarrow3x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow x=0;x=-5;x=5\)

(x+1)(x-2)(x-3)(2x-1)(3x-1)=0

\(x^3+3x^2+2x=0\)

\(\Leftrightarrow x\left(x^2+3x+2\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)

\(x^4+3x^3-x-3=0\)

\(\Leftrightarrow x^3\left(x+3\right)-\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^3-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^3-1=0\end{matrix}\right.\)            \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

Vậy \(x\in\left\{-3;1\right\}\)