Đồ thị hàm số 

A

- Hướng dẫn giải

Chọn A.

Gọi $x_1;x_2;x_3$ là 3 nghiệm phân biệt của PT $x^3-3m{x}^2+9x-7=0$

Áp dụng định lý Vi – ét cho PT bậc 3 có:

$\left\{\begin{array}{l}{x}_1+x_2+x_3=-\frac{b}{a}\\ x_1{x}_2+x_1{x}_3+x_2{x}_3=\frac{c}{a}\\ x_1{x}_2{x}_3=-\frac{d}{a}\end{array}\right.$ nên có $\left\{\begin{array}{l}{x}_1+x_2+x_3=-\frac{-3m}{1}=3m\\ x_1{x}_2+x_1{x}_3+x_2{x}_3=\frac{9}{1}=9\\ x_1{x}_2{x}_3=-\frac{7}{1}=7\end{array}\right.$

Để $x_1;x_2;x_3$ lập thành 1 cấp số cộng, ta giả sử $u_1=x_1,u_2=x_2;u_3=x_3$ tức là $x_2=x_1+d$ , $x_3=x_1=2d$

Khi đó ta có:

$\left\{\begin{array}{l}3x_1+3d=3m\\ x_1\left(x_1+d\right)+x_1\left(x_1+2d\right)+\left(x_1+d\right)\left(x_1+2d\right)=9\\ x_1\left(x_1+d\right)\left(x_1+2d\right)=7\end{array}\right.$

 $\iff \left\{\begin{array}{l}{x}_1=m-d\\ \left(m-d\right)\left(m-d+d\right)+\left(m-d\right)\left(m-d+2d\right)\\ +\left(m-d+d\right)\left(m-d+2d\right)=9\\ \left(m-d\right)\left(m-d+d\right)\left(m-d+2d\right)=7\end{array}\right.$

$\iff \left\{\begin{array}{l}{x}_1=m-d\\ \left(m-d\right)m+\left(m-d\right)\left(m+d\right)+m\left(m+d\right)=9\\ \left(m-d\right)m\left(m+d\right)=7\end{array}\right.$

$\iff \left\{\begin{array}{l}{x}_1=m-d\\ m^2-md+m^2+md+m^2-d^2=9\\ \left(m-d\right)m\left(m+d\right)=7\end{array}\right.$

$\iff \left\{\begin{array}{l}{x}_1=m-d\\ 3m^2-d^2=9\\ \left(m-d\right)m\left(m+d\right)=7\end{array}\right.$$\iff \left\{\begin{array}{l}{x}_1=m-d\\ d^2=3m^2-9\\ m\left(m^2-d^2\right)=7\end{array}\right.$

$\iff \left\{\begin{array}{l}{x}_1=m-d\\ d^2=3m^2-9\\ m\left(m^2-\left(3m^2-9\right)\right)=7\end{array}\right.$$\iff \left\{\begin{array}{l}{x}_1=m-d\\ d^2=3m^2-9\\ m\left(-2m^2+9\right)=7\end{array}\right.$

$\iff \left\{\begin{array}{l}{x}_1=m-d\\ d^2=3m^2+9\\ -2m^3+9m=7\end{array}\right.$$\iff \left[\begin{array}{l}m=1\\ m=\frac{-1+\sqrt{15}}{2}\\ m=\frac{-1-\sqrt{15}}{2}\end{array}\right.$