Question

có bao nhiêu số nguyeen m để pt \(2x^2-6x=\left(x+1\right)\sqrt{4x+m}+m\) có đúng 1 nghiệm

Answer 1

ĐKXĐ: ...

Đặt \(\sqrt{4x+m}=t\ge0\Rightarrow m=t^2-4x\)

\(2x^2-6x=\left(x+1\right)t+t^2-4x\)

\(\Leftrightarrow2x^2-x\left(t+2\right)-t^2-t=0\)

\(\Delta=\left(t+2\right)^2+8\left(t^2+t\right)=\left(3t+2\right)^2\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{t+2+3t+2}{4}=t+1\\x=\dfrac{t+2-3t-2}{4}=-\dfrac{t}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{4x+m}=x-1\left(x\ge1\right)\\\sqrt{4x+m}=-2x\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}m=x^2-6x+1\left(x\ge1\right)\\m=4x^2-4x\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}m=-8\\-4< m< 0\end{matrix}\right.\)