Question

Cho biểu thức:

P = (\(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\)):\(\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)

Tìm m để vs mọi giá trị x>9 ta có: m(\(\sqrt{x}\)-3)P>x+1

- Cô ơi giải giúp em với

Answer 1

Rút gọn P:

\(P=\dfrac{4x}{\sqrt{x}-3}\)

\(\Rightarrow m\left(\sqrt{x}-3\right)P>x+1\)

\(\Leftrightarrow4mx>x+1\)

\(\Leftrightarrow\left(4m-1\right)x>1\)(1)

Xét \(4m-1=0\)

\(\Leftrightarrow m=\dfrac{1}{4}\)(loại vì (1) sai)

\(\Leftrightarrow x>\dfrac{1}{4m-1}\)

Xét \(\dfrac{1}{4m-1}< 0\)

\(\Rightarrow\left\{{}\begin{matrix}x< 0\\x>9\end{matrix}\right.\)(loại)

Xét \(\Leftrightarrow\dfrac{1}{4m-1}>0\)

\(\Leftrightarrow\left\{{}\begin{matrix}4m-1>0\\\dfrac{1}{4m-1}\le9\end{matrix}\right.\)

\(\Leftrightarrow m\ge\dfrac{5}{18}\)

Answer 2

ta có : \(P=\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)

\(P=\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\dfrac{1-\sqrt{x}}{\sqrt{x}\left(2-\sqrt{x}\right)}-\dfrac{2}{\sqrt{x}}\right)\)

\(P=\left(\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\dfrac{1-\sqrt{x}-2\left(2-\sqrt{x}\right)}{\sqrt{x}\left(2-\sqrt{x}\right)}\right)\)

\(P=\left(\dfrac{8\sqrt{x}-4x+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\dfrac{1-\sqrt{x}-4+2\sqrt{x}}{\sqrt{x}\left(2-\sqrt{x}\right)}\right)\) \(P=\left(\dfrac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\dfrac{\sqrt{x}-3}{\sqrt{x}\left(2-\sqrt{x}\right)}\right)\) \(P=\left(\dfrac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right).\left(\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\right)\)

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

\(\Rightarrow m\left(\sqrt{x}-3\right)P>x+1\Leftrightarrow4mx>x+1\Leftrightarrow\left(4m-1\right)x>1\) (1)

th1: \(m=\dfrac{1}{4}\) \(\Rightarrow\) loại vì (1) vô nghĩa

th2: \(m>\dfrac{1}{4}\) \(\Rightarrow x>\dfrac{1}{4m-1}\) vì \(x>9\)

\(\Rightarrow\) để \(x>9\) là điều chắc chắn thì \(\dfrac{1}{4m-1}\ge9\Leftrightarrow1\ge36m-9\Leftrightarrow m\le\dfrac{5}{18}\)

\(\Rightarrow\dfrac{1}{4}< m\le\dfrac{5}{18}\)

th3: \(m< \dfrac{1}{4}\) \(\Rightarrow x< \dfrac{1}{4m-1}\) mà \(x>9\) \(\Rightarrow\dfrac{1}{4m-1}>9\) \(\Leftrightarrow\) \(m< \dfrac{5}{18}\)

\(\Rightarrow m< \dfrac{1}{4}\)

vậy \(\dfrac{1}{4}< m\le\dfrac{5}{18}\) hoặc \(m< \dfrac{1}{4}\)