Bài 3:
a: Để hàm số đồng biến thì (k-6)(k+1)>0
\(\Leftrightarrow\left[{}\begin{matrix}k>6\\k< -1\end{matrix}\right.\)
b: Để hàm số nghịch biến thì (k-6)(k+1)<0
hay -1<k<6
1) a) ĐKXĐ : \(x>0\)
b) \(A=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+1\)
\(=\dfrac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+1\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-2\sqrt{x}-1+1\)
\(=\sqrt{x}\left(\sqrt{x}+1\right)-2\sqrt{x}=x-\sqrt{x}\)
c) Ta có : \(x-\sqrt{x}=\sqrt{x}\left(\sqrt{x}-1\right)>0\)(Vì x > 1 => \(\sqrt{x}>1\))
=> \(\left|x-\sqrt{x}\right|=x-\sqrt{x}=A\)
=> |A| = A
d) Khi A = 2
=> \(x-\sqrt{x}=2\)
<=> \(\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=0\)
<=> \(\left[{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{x}+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x\in\varnothing\end{matrix}\right.\Leftrightarrow x=4\)
Vậy khi x = 4 thì A = 2
d) A = \(x-\sqrt{x}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
=> Min A = 1/4 khi x = 1/4
