Question

cho pt: x² - 2mx + 4m -3 = 0 (1)

tìm m để pt có 2 nghiệm phân biệt x1, x2 sao cho

a) x₁ > 1 > x₂

b) x₁ > 2, x₂ > 2

Answer 1

\(\Delta'=m^2-4m+3>0\Rightarrow\left[{}\begin{matrix}m< 1\\m>3\end{matrix}\right.\)

Đặt \(f\left(x\right)=x^2-2mx+4m-3\)

a/ Để \(x_2< 1< x_1\)

\(\Leftrightarrow f\left(1\right)< 0\Leftrightarrow1-2m+4m-3< 0\)

\(\Leftrightarrow2m< 2\Rightarrow m< 1\)

b/ Để \(x_1>2;x_2>2\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x_1-2\right)\left(x_2-2\right)>0\\\frac{x_1+x_2}{2}>2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1x_2-2\left(x_1+x_2\right)+4>0\\x_1+x_2>4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4m-3-4m+4>0\\2m>4\end{matrix}\right.\) \(\Rightarrow m>2\)

\(\Rightarrow m>3\)