Question

Bài 1: Cho pt \(x^2-4x+m=2\sqrt{4x-x^2}\) . Tìm m để phương trình có nghiệm

Bài 2: Cho pt : \(x^2+4x+m+3+\sqrt{x^2+4x+5}=0\)

Tìm m để pt có nghiệm |x| nhỏ hơn hoặc bằng 5

Answer 1

1)Dat t=\(\sqrt{4x-x^2}\)\(\Rightarrow Pt\Leftrightarrow t^2+2t+1=m+1\ge0\Rightarrow m\ge-1\)

Theo dinh li Viet thi \(\left\{{}\begin{matrix}t_1+t_2=-2\\t_1t_2=-m\end{matrix}\right.\Rightarrow-m\le0\Leftrightarrow m\ge0}\)

Answer 2

Dat \(t=\sqrt{x^2+4x+5}\left(t\ge1\right)\)\(\Rightarrow Pt\Leftrightarrow t^2+t+m-2=0\)

DK:\(\Delta=1-4\left(m-2\right)=9-4m\ge0\Leftrightarrow m\le\dfrac{9}{4}\)

Pt co nghiem la \(t=\dfrac{-1-\sqrt{\Delta}}{2}\left(loai\right),t=\dfrac{-1+\sqrt{\Delta}}{2}\)

Vi \(t\ge1\)\(\Rightarrow\sqrt{\Delta}\ge3\Leftrightarrow9-4m\ge9\Leftrightarrow m\le0\)

\(5\ge\left|x\right|=\left|\sqrt{\dfrac{-1+\sqrt{9-4m}}{2}}\right|=\sqrt{\dfrac{-1+\sqrt{9-4m}}{2}}\Leftrightarrow\sqrt{9-4m}\le51\Leftrightarrow m\ge-648\)Vay \(-648\le m\le0\)