Bài 3
a) \(x^2-5x+m-2=0\)
Thay \(m=-4\) vào phương trình
\(\Rightarrow x^2-5x-6=0\)
\(\Delta=b^2-4ac\)
\(\Delta=49\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{5+\sqrt{49}}{2}=6\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{5-\sqrt{49}}{2}=-1\end{matrix}\right.\)
b )
\(x^2-5x+m-2=0\)
\(\Delta=b^2-4ac\)
\(\Delta=33-4m\)
Theo định lý Viet
\(\Rightarrow\left\{{}\begin{matrix}P=x_1+x_2=\dfrac{-b}{a}\\S=x_1x_2=\dfrac{c}{a}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}P=x_1+x_2=5\\S=x_1x_2=m-2\end{matrix}\right.\)
Để phương trình có 2 nghiệm dương phân biệt
\(\Rightarrow\left\{{}\begin{matrix}\Delta>0\\P>0\\S>0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}33-4m>0\\m-2>0\\5>0\left(đúng\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m< \dfrac{33}{4}\\m>2\end{matrix}\right.\)
\(\Rightarrow2< m< \dfrac{33}{4}\)
Ta có \(2\left(\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}\right)=3\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}=\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x_1}+\sqrt{x_2}}{\sqrt{x_1x_2}}=\dfrac{3}{2}\)
\(\Leftrightarrow\left(\dfrac{\sqrt{x_1}+\sqrt{x_2}}{\sqrt{x_1x_2}}\right)^2=\dfrac{9}{4}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x_1}+\sqrt{x_2}\right)^2}{x_1x_2}=\dfrac{9}{4}\)
\(\Leftrightarrow\dfrac{x_1+x_2+2\sqrt{x_1x_2}}{x_1x_2}=\dfrac{9}{4}\)
\(\Leftrightarrow\dfrac{5+\sqrt{m-2}}{m-2}=\dfrac{9}{4}\)
\(\Leftrightarrow20+4\sqrt{m-2}=9m-18\)
\(\Leftrightarrow4\sqrt{m-2}=9m-38\)
\(\Leftrightarrow64m-128=\left(9m-38\right)^2\)
\(\Leftrightarrow64m-128=81m^2-684m+1444\)
\(\Leftrightarrow81m^2-748m+1572=0\)
\(\Delta=b^2-4ac\)
\(\Delta=50176\)
\(\Rightarrow\left\{{}\begin{matrix}m_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{748+\sqrt{50176}}{162}=6\\m_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{748-\sqrt{50176}}{162}=\dfrac{262}{81}\end{matrix}\right.\)
Vì \(2< m< \dfrac{33}{4}\)
\(\Rightarrow m\in\left\{6;\dfrac{262}{81}\right\}\)
