a) `(3x-2)(x+6)(x^2+5)=0`
`=>` \(\left[ \begin{array}{l}3x-2=0\\x+6=0\\x^2+5=0\end{array} \right.\) `=>` \(\left[ \begin{array}{l}x=\dfrac{2}{3}\\x=-6\\x^2=-5(\text{loại})\end{array} \right.\)
Vậy `S={-6;2/3}`
b) `(2x+5)^2=4(3x-1)^2`
`=>(2x+5)^2=[2(3x-1)]^2`
`=>(2x+5)^2=(6x-2)^2`
`=>(2x+5)^2-(6x-2)^2=0`
`=>(2x+5-6x+2)(2x+5+6x-2)=0`
`=>(7-4x)(8x+3)=0`
`=>` \(\left[ \begin{array}{l}7-4x=0\\8x+3=0\end{array} \right.\) `=>` \(\left[ \begin{array}{l}x=\dfrac{7}{4}\\x=-\dfrac{3}{8}\end{array} \right.\)
Vậy `S={-3/8;7/4}`
c) `4x^2(x-1)-x+1=0`
`=>4x^2(x-1)-(x-1)=0`
`=>(4x^2-1)(x-1)=0`
`=>` \(\left[ \begin{array}{l}4x^2-1=0\\x-1=0\end{array} \right.\) `=>` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\\x=1\end{array} \right.\)
Vậy `S={+-1/2;1}`
d) `(2x+5)(x-4)=(x-4)(5-x)`
`=>(2x+5)(x-4)-(x-4)(5-x)=0`
`=>(x-4)(2x+5-5+x)=0`
`=>3x(x-4)=0`
`=>` \(\left[ \begin{array}{l}3x=0\\x-4=0\end{array} \right.\) `=>` \(\left[ \begin{array}{l}x=0\\x=4\end{array} \right.\)
Vậy `S={0;4}`
