`a)` `(3x-2)(4x+5)=0`
`<=>` \(\left[ \begin{array}{l}3x-2=0\\4x+5=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{array} \right.\)
Vậy `x∈{2/3;-5/4}
`b)` `(x^2-4)(x-2)(3-1x)=0`
`<=>` `(x^2-4)(3-1x)=0` ( ko cần lấy `x-2` vì `x^2-4` đã bao hàm nó rồi )
`<=>` \(\left[ \begin{array}{l}x^2-4=0\\3-1x=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=±2\\x=3\end{array} \right.\)
Vậy `x∈{2;-2;3}`
`c)` `(2x-5)^2-(x+2)^2=0`
`<=>` (2x-5-x-2)(2x-5-x+2)=0`
`<=>` (x-7)(x-3)=0`
`<=>` \(\left[ \begin{array}{l}x-7=0\\x-3=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=7\\x=3\end{array} \right.\)
Vậy `x∈{7;3}`
`d)` `x(2x-9)=3x(x-5)`
`⇔2x^2-9x=3x^2-15x`
`⇔2x^2-9x-3x^2+15x=0`
`⇔-x^2+6x=0`
`⇔-x(x-6)=0`
`⇔x(x-6)=0`
`<=>` \(\left[ \begin{array}{l}x=0\\x-6=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=0\\x=6\end{array} \right.\)
Vậy `x∈{0;6}`
