Giải thích các bước giải:
$a) 2x² -5x +3 = 0$
$⇔ 2x.(x -1) -3.(x -1) = 0$
$⇔ (x -1).(2x -3) = 0$
$⇔ \left[ \begin{array}{l}x -1=0\\2x -3=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=1\\x=\dfrac{3}{2}\end{array} \right.$
Vậy `S = {1; 3/2}`
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$b) 4x² +6x -1 = 0$
`⇔ (2x +3/2)² -13/4 = 0`
$⇔ (2x +\dfrac{3}{2} -\dfrac{\sqrt{13}}{2}).(2x +\dfrac{3}{2} +\dfrac{\sqrt{13}}{2}) = 0$
$⇔ \left[ \begin{array}{l}2x +\dfrac{3}{2} -\dfrac{\sqrt{13}}{2}=0\\2x +\dfrac{3}{2} +\dfrac{\sqrt{13}}{2}=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=\dfrac{-3 +\sqrt{13}}{4}\\x=\dfrac{-3 -\sqrt{13}}{4}\end{array} \right.$
Vậy `S = {`$\dfrac{-3 +\sqrt{13}}{4}; \dfrac{-3 -\sqrt{13}}{4}$`}`
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$c) 2x² +x -1 = 0$
$⇔ 2x.(x +1) -(x +1) = 0$
$⇔ (x +1).(2x -1) = 0$
$⇔ \left[ \begin{array}{l}x +1=0\\2x -1=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=-1\\x=\dfrac{1}{2}\end{array} \right.$
Vậy `S = {-1; 1/2}`
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$d) 3x² +2x -1 = 0$
$⇔ 3x.(x +1) -(x +1) = 0$
$⇔ (x +1).(3x -1) = 0$
$⇔ \left[ \begin{array}{l}x +1=0\\3x -1=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=-1\\x=\dfrac{1}{3}\end{array} \right.$
Vậy `S = {-1; 1/3}`
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$e) x² -x +1 = 0$
`⇔ (x -1/2)² +3/4 = 0` (Vô lý)
Vì `(x -1/2)² +3/4 > 0` (vs ∀ x)
Vậy `S = ∅`
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$g) x² +x +1 = 0$
`⇔ (x +1/2)² +3/4 = 0` (Vô lý)
Vì `(x +1/2)² +3/4 > 0` (vs ∀ x)
Vậy `S = ∅`
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$h) 2x² -5x +7 = 0$
`⇔ 2.(x² -5/2x +7/2) = 0`
`⇔ 2.[(x -5/4)² +31/16] = 0`
`⇔ 2.(x -5/4)² +31/8 = 0` (Vô lý)
Vì `2.(x -5/4)² +31/8 > 0` (vs ∀ x)
Vậy `S = ∅`
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$k) 4x² -7x +3 = 0$
$⇔ 4x.(x -1) -3.(x -1) = 0$
$⇔ (x -1).(4x -3) = 0$
$⇔ \left[ \begin{array}{l}x -1=0\\4x -3=0\end{array} \right. ⇔ \left[ \begin{array}{l}x=1\\x=\dfrac{3}{4}\end{array} \right.$
Vậy `S = {1; 3/4}`
