a: \(\Leftrightarrow x^2+6x+9-6x+3>x^2-4x\)
=>-4x<12
hay x>-3
b: \(\Leftrightarrow6+2\left(x+1\right)>2x-1-12\)
=>6+2x+2>2x-13
=>8>-13(luôn đúng)
c: \(\Leftrightarrow12x-3\left(2x-7\right)< 8x-6\left(2x+3\right)-12\)
=>12x-6x+21<8x-12x-18-12
=>6x+21<-4x-30
=>10x<-51
hay x<-51/10
h*) \(\left(2x-1\right)\left(3-2x\right)\left(1-x\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1>0\\3-2x>0\\1-x>0\end{matrix}\right.\) ; \(\left[{}\begin{matrix}2x-1< 0\\3-2x< 0\\1-x>0\end{matrix}\right.\) ; \(\left[{}\begin{matrix}2x-1< 0\\3-2x>0\\1-x< 0\end{matrix}\right.\) ; \(\left[{}\begin{matrix}2x-1>0\\3-2x< 0\\1-x< 0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{2}\\x< \dfrac{3}{2}\\x< 1\end{matrix}\right.\)(có thể xảy ra) ; \(\left[{}\begin{matrix}x< \dfrac{1}{2}\\x>\dfrac{3}{2}\\x< 1\end{matrix}\right.\) (vô lí) ; \(\left[{}\begin{matrix}x< \dfrac{1}{2}\\x< \dfrac{3}{2}\\x>1\end{matrix}\right.\)(vô lí) ; \(\left[{}\begin{matrix}x>\dfrac{1}{2}\\x>\dfrac{3}{2}\\x>1\end{matrix}\right.\) (có thể xảy ra)
