Question

Cho biểu thức \(P=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)

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

(KQ: \(P=\frac{4x}{\sqrt{x}-3}\)

Answer 1

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

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

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

\(P=\frac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-5x\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)

\(P=\frac{4\sqrt{x}\left(2+5x\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)

\(P=\frac{4\sqrt{x}}{2-\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)

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

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

Có:

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

\(\Leftrightarrow m\left(\sqrt{x}-3\right).\frac{4x}{\sqrt{x}-3}>x+1\)

\(\Leftrightarrow4mx>x+1\)

\(\Leftrightarrow4mx-x>1\)

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

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

Lại có:

\(x>9\)

\(\Rightarrow\frac{1}{4m-1}< 9\)

\(\Leftrightarrow1< 9\left(4m-1\right)\)

\(\Leftrightarrow1< 36m-1\)

\(\Leftrightarrow10< 36m\)

\(\Leftrightarrow m< \frac{5}{18}\)

Answer 2

Ấy, nhầm nha. 

Đoạn cuối là m<5/18

Vội quá gõ nhầm. 

Answer 3

Ấy, nhầm nha. 

Đoạn cuối là m<5/18

Vội quá gõ nhầm.