Đáp án:
Giải thích các bước giải:
b) `x^2-3x-7=0`
`⇔ x^2 - 2 . 3/2 x+9/4-37/4=0`
`⇔ (x-3/2)^2-(\frac{\sqrt{37}}{2})^2=0`
`⇔ (x-3/2-\frac{\sqrt{37}}{2})(x-3/2+\frac{\sqrt{37}}{2})=0`
`⇔` \(\left[ \begin{array}{l}x-\dfrac{3}{2}-\dfrac{\sqrt{37}}{2}=0\\x-\dfrac{3}{2}+\dfrac{\sqrt{37}}{2}=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{3}{2}+\dfrac{\sqrt{37}}{2}\\x=\dfrac{3}{2}-\dfrac{\sqrt{37}}{2}\end{array} \right.\)
Vậy `S={\frac{3}{2}+\frac{\sqrt{37}}{2};\frac{3}{2}-\frac{\sqrt{37}}{2}}`
c) `\frac{x-1}{2}.\frac{x-2}{3} \le x-\frac{x-3}{4}`
`⇔ \frac{(x-1)(x-2)}{6} - x+\frac{x-3}{4} \le 0`
`⇔ \frac{2(x-1)(x-2)}{12} - \frac{12x}{12}+\frac{3(x-3)}{12} \le 0`
`⇔ \frac{2(x^2-3x+2)-12x+3(x-3)}{12} \le 0`
`⇔ \frac{2x^2-15x-5}{12} \le 0`
`⇔ 2x^2-15x-5 \le 0`
`⇔ \frac{15-\sqrt{265}}{4} \le x \le \frac{15+\sqrt{265}}{4}`
Vậy `S={x|\frac{15-\sqrt{265}}{4} \le x \le \frac{15+\sqrt{265}}{4}}`
