Question

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

a) Tìm giá trị của x để P xác định

b) Rút gọn P

c) Tìm x sao cho P>1

Answer 1

ĐKXĐ: \(x>0;x\ne4;9\)

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

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

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

\(P>1\Rightarrow\frac{4x}{\sqrt{x}-3}>1\Rightarrow\frac{4x}{\sqrt{x}-3}-1>0\)

\(\Rightarrow\frac{4x-\sqrt{x}+3}{\sqrt{x}-3}>0\Rightarrow\sqrt{x}-3>0\Rightarrow x>9\)