Question

\(\sqrt{9-x}=\sqrt{3m-x^2+9x}-\sqrt{x}\)

Answer 1

ĐK: \(0\le x\le9\)

\(pt\Leftrightarrow\sqrt{9-x}+\sqrt{x}=\sqrt{3m-x^2+9x}\)

\(\Leftrightarrow9+2\sqrt{9x-x^2}=3m-x^2+9x\)

\(\Leftrightarrow3m=x^2-9x+2\sqrt{9x-x^2}+9\left(1\right)\)

Đặt \(\sqrt{9x-x^2}=t\left(0\le t\le\dfrac{9}{2}\right)\)

\(\left(1\right)\Leftrightarrow3m=f\left(t\right)=-t^2+2t+9\)

Yêu cầu bài toán thỏa mãn khi 

\(minf\left(t\right)\le3m\le maxf\left(t\right)\)

\(\Leftrightarrow-\dfrac{9}{4}\le3m\le10\)

\(\Leftrightarrow-\dfrac{3}{4}\le m\le\dfrac{10}{3}\)