Giải thích các bước giải:
Cách giải: `A(x).B(x)=0⇔`\(\left[ \begin{array}{l}A(x)=0\\B(x)=0\end{array} \right.\)
a, `(5x-4)(4x+6)=0`
`⇔`\(\left[ \begin{array}{l}5x-4=0\\4x+6=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\dfrac{4}{5}\\x=-\dfrac{2}{3}\end{array} \right.\)
Vậy `S={\frac{4}{5};-\frac{2}{3}}`
b, `(4x-10)(24+5x)=0`
`⇔`\(\left[ \begin{array}{l}4x-10=0\\24+5x=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\dfrac{5}{2}\\x=-\dfrac{24}{5}\end{array} \right.\)
Vậy `S={\frac{4}{2};-\frac{24}{5}}`
c, `(5x-10)(8-2x)=0`
`⇔`\(\left[ \begin{array}{l}5x-10=0\\8-2x=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=2\\x=4\end{array} \right.\)
Vậy `S={2;4}`
d, `(3,5x-7)(2,1x-6)=0`
`⇔`\(\left[ \begin{array}{l}3,5x-7=0\\2,1x-6=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=2\\x=\dfrac{20}{7}\end{array} \right.\)
Vậy `S={2;\frac{20}{7}}`
e, `(9-3x)(15+3x)=0`
`⇔`\(\left[ \begin{array}{l}9-3x=0\\15+3x=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=3\\x=-5\end{array} \right.\)
Vậy `S={3;-5}`
f, `(x-3)(2x+1)=0`
`⇔`\(\left[ \begin{array}{l}x-3=0\\2x+1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=3\\x=-\dfrac{1}{2}\end{array} \right.\)
Vậy `S={3;-\frac{1}{2}}`
