Đáp án:
`a) 2x² - 7x + 3 = 0`
`⇔ 2x² - 6x - x + 3 = 0`
`⇔ 2x.(x - 3) - (x - 3)=0`
`⇔ (2x - 1).(x - 3) = 0`
⇔ \(\left[ \begin{array}{l}2x-1=0\\x-3=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{1}{2}\\x=3\end{array} \right.\)
Vậy `S = {\frac{1}{2};3}`
`b) 10x^{2}-29x+10=0`
`⇔ 10x² - 25x - 4x + 10 = 0`
`⇔ 5x.(2x - 5) - 2(2x - 5)=0`
`⇔ (5x - 2).(2x - 5) = 0`
⇔ \(\left[ \begin{array}{l}5x-2=0\\2x-5=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{2}{5}\\x=\frac{5}{2}\end{array} \right.\)
Vậy `S = {\frac{2}{5};\frac{5}{2}}`
`c) 6x^{2}-13x+6=0`
`⇔ 6x² - 4x - 9x + 6 = 0`
`⇔ 2x.(3x - 2) - 3(3x - 2)=0`
`⇔ (2x - 3).(3x - 2) = 0`
⇔ \(\left[ \begin{array}{l}2x-3=0\\3x-2=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{3}{2}\\x=\frac{2}{3}\end{array} \right.\)
Vậy `S = {\frac{3}{2};\frac{2}{3}}`
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