`a)x(x-1)=0`
`⇒`\(\left[ \begin{array}{l}x=0\\x-1=0\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\)
Vậy `x∈{0;1}`
`b)3x^2-6x=0`
`⇒3x(x-2)=0`
`⇒`\(\left[ \begin{array}{l}3x=0\\x-2=0\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x=0\\x=2\end{array} \right.\)
Vậy `x∈{0;2}`
`c)x(x-6)+10(x-6)=0`
`⇒(x-6)(x+10)=0`
`⇒`\(\left[ \begin{array}{l}x-6=0\\x+10=0\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x=6\\x=-10\end{array} \right.\)
Vậy `x∈{6;-10}`
`d)4x(2x-3)+5(3-2x)=0`
`⇒4x(2x-3)-5(2x-3)=0`
`⇒(2x-3)(4x-5)=0`
`⇒`\(\left[ \begin{array}{l}2x-3=0\\4x-5=0\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}2x=3\\4x=5\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=\dfrac{5}{4}\end{array} \right.\)
Vậy `x∈{3/2;5/4}`
`e)(3x-5)(2x-4)=9x^2-25`
`⇒(3x-5)(2x-4)-[(3x)^2-5^2]=0`
`⇒(3x-5)(2x-4)-(3x-5)(3x+5)=0`
`⇒(3x-5)[2x-4-(3x+5)]=0`
`⇒(3x-5)(2x-4-3x-5)=0`
`⇒(3x-5)(-x-9)=0`
`⇒`\(\left[ \begin{array}{l}3x-5=0\\-x-9=0\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}3x=5\\-x=9\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x=\dfrac{5}{3}\\x=-9\end{array} \right.\)
Vậy `x∈{5/3;-9}.`
