Bài 1:
a) `x^2-5x+1=0`
`⇔ x^2-5x+\frac{25}{4}-\frac{21}{4}=0`
`⇔ (x-\frac{5}{2})^2-\frac{21}{4}=0`
`⇔ (x-\frac{5}{2})^2-(\frac{\sqrt{21}}{2})^2=0`
`⇔ (x-\frac{5}{2}+\frac{\sqrt{21}}{2})(x-\frac{5}{2}-\frac{\sqrt{21}}{2})=0`
`⇔` \(\left[ \begin{array}{l}x-\dfrac{5}{2}+\dfrac{\sqrt{21}}{2}=0\\x-\dfrac{5}{2}-\dfrac{\sqrt{21}}{2}=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{5-\sqrt{21}}{2}\\x=\dfrac{5+\sqrt{21}}{2}\end{array} \right.\)
Vậy `S={\frac{5-\sqrt{21}}{2};\frac{5+\sqrt{21}}{2}}`
b) `3x^2-12x-1=0`
`⇔ (\sqrt{3}x-2\sqrt{3})^2-13=0`
`⇔ (\sqrt{3}x-2\sqrt{3})^2-(\sqrt{13})^2=0`
`⇔ (\sqrt{3}x-2\sqrt{3}+\sqrt{13}).(\sqrt{3}x-2\sqrt{3}-\sqrt{13})=0`
`⇔` \(\left[ \begin{array}{l}\sqrt{3}x-2\sqrt{3}+\sqrt{13}=0\\\sqrt{3}x-2\sqrt{3}-\sqrt{13}=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{6+\sqrt{39}}{3}\\x=\dfrac{6-\sqrt{39}}{3}\end{array} \right.\)
Vậy `S={\frac{6+\sqrt{39}}{3};\frac{6-\sqrt{39}}{3}}`
Bài 2:
$a)A=1/4x^2-x+1$
$A=1/4x^2-2×1/2x+1$
$A=(1/2x+1)^2$
Ta có $(1/2x+1)^2≥0$
$⇒A=(1/2x+1)^2≥0$
⇒$A_{min}=0$ $⇔x=-2$
$b)B=3x^2-4x-2$
$B=2x^2+(x^2-4x+4)-6$
$B=2x^2+(x-2)^2-6$
Ta có $:2x^2≥0$
$(x-2)^2≥0$
$⇒B=2x^2+(x-2)^2-6≥-6$
$B_{min}=-6$
