Question

cíu tui

Answer 1

a. Em tự giải

b.

\(B=\dfrac{3\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{x+6}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{3\left(\sqrt{x}-2\right)+\sqrt{x}\left(\sqrt{x}-4\right)+x+6}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}\)

c.

\(A< B\Rightarrow\dfrac{x}{\sqrt{x}-2}< \dfrac{2\sqrt{x}-1}{\sqrt{x}-2}\)

\(\Leftrightarrow\dfrac{x}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}< 0\)

\(\Leftrightarrow\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-2}< 0\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-2}< 0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-1\ne0\\\sqrt{x}-2< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}\ne1\\\sqrt{x}< 2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x< 4\end{matrix}\right.\)

Kết hợp ĐKXĐ \(\Rightarrow\left\{{}\begin{matrix}0< x< 4\\x\ne1\end{matrix}\right.\)