$3x^2+5x-6=0\\\text{Ta có: }\Delta=b^2-4ac\\\hspace{1,8cm}=5^2-4.3.(-6)\\\hspace{1,8cm}=97>0\\\to \text{Phương trình có 2 nghiệm phân biệt}\\\to \text{Theo Viet: }\begin{cases}x_1+x_2=-\dfrac{b}{a}=-\dfrac{5}{3}\\x_1x_2=\dfrac{c}{a}=-2\end{cases}\\\text{Theo đề bài: }A=\dfrac{x_1}{x_2-1}+\dfrac{x_2}{x_1-1}\\\hspace{3cm}=\dfrac{x_1(x_1-1)+x_2(x_2-1)}{(x_2-1)(x_1-1)}\\\hspace{3cm}=\dfrac{x_1^2-x_1+x_2^2-x_2}{x_1x_2-x_1-x_2+1}\\\hspace{3cm}=\dfrac{(x_1+x_2)^2-2x_1x_2-(x_1+x_2)}{x_1x_2-(x_1+x_2)+1}\\\hspace{3cm}=\dfrac{\Big(-\dfrac{5}{3}\Big)^2-2.(-2)-\Big(-\dfrac{5}{3}\Big)}{-2-\Big(-\dfrac{5}{3}\Big)+1}\\\hspace{3cm}=\dfrac{38}{3}$
Đáp án:
3x² + 5x – 6 = 0
⇒ \(x_1\) = $\frac{-5+\sqrt{97}}{6}$ , \(x_2\) = $\frac{-5-\sqrt{97}}{6}$
A = $\frac{x_1}{x_2-1}$ + $\frac{x_2}{x_1-1}$
A = $\frac{\frac{-5+\sqrt{97}}{6}}{\frac{-5-\sqrt{97}}{6}-1}$ + $\frac{\frac{-5-\sqrt{97}}{6}}{\frac{-5+\sqrt{97}}{6}-1}$
A = $\frac{55-11\sqrt{97}}{12}$
Giải thích các bước giải:
$3x^2+5x-6=0\\\text{Ta có: }\Delta=b^2-4ac\\\hspace{1,8cm}=5^2-4.3.(-6)\\\hspace{1,8cm}=97>0\\\to \text{Phương trình có 2 nghiệm phân biệt}\\\to \text{Theo Viet: }\begin{cases}x_1+x_2=-\dfrac{b}{a}=-\dfrac{5}{3}\\x_1x_2=\dfrac{c}{a}=-2\end{cases}\\\text{Theo đề bài: }A=\dfrac{x_1}{x_2-1}+\dfrac{x_2}{x_1-1}\\\hspace{3cm}=\dfrac{x_1(x_1-1)+x_2(x_2-1)}{(x_2-1)(x_1-1)}\\\hspace{3cm}=\dfrac{x_1^2-x_1+x_2^2-x_2}{x_1x_2-x_1-x_2+1}\\\hspace{3cm}=\dfrac{(x_1+x_2)^2-2x_1x_2-(x_1+x_2)}{x_1x_2-(x_1+x_2)+1}\\\hspace{3cm}=\dfrac{\Big(-\dfrac{5}{3}\Big)^2-2.(-2)-\Big(-\dfrac{5}{3}\Big)}{-2-\Big(-\dfrac{5}{3}\Big)+1}\\\hspace{3cm}=\dfrac{38}{3}$