Question

Cho phương trình \(x^2-3mx-m=0\) có 2 nghiệm \(x_1,x_2\)

Tìm GTNN \(A=\dfrac{m^2}{x_2^2+3mx_1+3m}+\dfrac{x_1^2+3mx_2+3m}{m^2}\)

Answer 1

\(x^2-3mx-m=0\)

Theo định lý Viet

\(\Rightarrow\left\{{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}\\x_1x_2=\dfrac{c}{a}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2=3m\\x_1x_2=-m\end{matrix}\right.\)

Ta có \(A=\dfrac{m^2}{x_2^2+3mx_1+3m}+\dfrac{x^2_1+3mx_2+3m}{m^2}\)

\(\Leftrightarrow A=\dfrac{m^2}{x^2_2+\left(x_1+x_2\right)x_1+3m}+\dfrac{x^2_1+\left(x_1+x_2\right)x_2+3m}{m^2}\)

\(\Leftrightarrow A=\dfrac{m^2}{x^2_2+x^2_1+x_1x_2+3m}+\dfrac{x^2_1+x^2_2+x_1x_2+3m}{m^2}\)

\(\Leftrightarrow A=\dfrac{m^2}{\left(x_1+x_2\right)^2-2x_1x_2+x_1x_2+3m}+\dfrac{\left(x_1+x_2\right)^2-2x_1x_2+x_1x_2+3m}{m^2}\)

\(\Leftrightarrow A=\dfrac{m^2}{\left(3m\right)^2-2\left(-m\right)+\left(-m\right)+3m}+\dfrac{\left(3m\right)^2-2\left(-m\right)+\left(-m\right)+3m}{m^2}\)

\(\Leftrightarrow A=\dfrac{m^2}{9m^2+4m}+\dfrac{9m^2+4m}{m^2}\)

\(\Leftrightarrow A=\dfrac{m^2}{m\left(9m+4\right)}+\dfrac{m\left(9m+4\right)}{m^2}\)

\(\Leftrightarrow A=\dfrac{m}{9m+4}+\dfrac{9m+4}{m}\)

\(\Leftrightarrow A=\dfrac{m^2+\left(9m+4\right)^2}{\left(9m+4\right)m}\)

Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm

\(\Rightarrow m^2+\left(9m+4\right)^2\ge2\sqrt{m^2\left(9m+4\right)^2}\)

\(\Rightarrow m^2+\left(9m+4\right)^2\ge2\left(9m+4\right)m\)

\(\Rightarrow\dfrac{m^2+\left(9m+4\right)^2}{\left(9m+4\right)m}\ge2\)

\(\Rightarrow A\ge2\)

Vậy \(A_{min}=2\)

Answer 2

Rút gọn A: ĐK tồn tại A: \(\left\{{}\begin{matrix}m\ne0\\9m^2+4m\ge0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m>0\\m\le-\dfrac{4}{9}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x^2_1+3mx_2+3m=3m\left(x_1+x_2\right)+4m\\x^2_2+3mx_1+3m=3m\left(x_1+x_2\right)+4m\\A=\dfrac{m}{3\left(x_1+x_2\right)+4}+\dfrac{3\left(x_1+x_2\right)+4}{m}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x_1=\dfrac{3m-\sqrt{9m^2+4m}}{2}\\x_2=\dfrac{3m+\sqrt{9m^2+4m}}{2}\end{matrix}\right.\) \(\Rightarrow x_1+x_2=3m\)

Thay vào biểu thức A

\(A=\dfrac{m}{9m+4}+\dfrac{9m+4}{m}=t+\dfrac{1}{t}\)

cần bổ xung đk cho A =>\(\left[{}\begin{matrix}m>0\\m< \dfrac{-4}{9}\end{matrix}\right.\) (*)

Hiển nhiên khi m> 0; giá trị A lớn--> đang tìm giá trị nhỏ nhất

xét khi m<-4/9 có A=2 khi m=-1/2 <-4/9 nhận

Đáp số : m=-1/2