Đáp án:
Giải thích các bước giải:
a) `4x^2+12x-7=0`
`⇔ 4x^2+12x+9-16=0`
`⇔ (2x+3)^2-(4)^2=0`
`⇔ (2x+3-4)(2x+3+4)=0`
`⇔ (2x-1)(2x+7)=0`
`⇔` \(\left[ \begin{array}{l}2x-1=0\\2x+7=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-\dfrac{7}{2}\end{array} \right.\)
Vậy `S={1/2;-7/2}`
b) `9x^2-24x+19=0`
`⇔ 9x^2-24x+16+3=0`
`⇔ (3x-4)^2+3=0`
Ta có: `(3x-4)^2 \ge 0 \forall x`
`⇒ (3x-4)^2+3 \ge 3 \forall x`
Vậy PT vô nghiệm
c) `x^2-3x-28=0`
`⇔ x^2+4x-7x-28=0`
`⇔ x(x+4)-7(x+4)=0`
`⇔ (x-7)(x+4)=0`
`⇔` \(\left[ \begin{array}{l}x-7=0\\x+4=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=7\\x=-4\end{array} \right.\)
Vậy `S={7;-4}`
d) `2x^2-7x+5=0`
`⇔ 2x^2-2x-5x+5=0`
`⇔ 2x(x-1)-5(x-1)=0`
`⇔ (2x-5)(x-1)=0`
`⇔` \(\left[ \begin{array}{l}2x-5=0\\x-1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{5}{2}\\x=1\end{array} \right.\)
Vậy `S={5/2;1}`
