Question

Cho biểu thức:

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

a) Rút gọn P

b) Tính P với \(x=3-2\sqrt{2}\)

c) Tìm x để P.\(\sqrt{x}\) nhận giá trị lớn nhất

Answer 1

a) điều kiện xác định : \(x\ge0;x\ne4;x\ne0\)

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

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

b) thay \(x=3-2\sqrt{2}\) vào \(P\) ta có : \(P=-\sqrt{3-2\sqrt{2}}-1\)

\(=-\sqrt{\left(\sqrt{2}-1\right)^2}-1=-\sqrt{2}+1-1=-\sqrt{2}\)

c) ta có : \(P\sqrt{x}=\sqrt{x}\left(-\sqrt{x}-1\right)=-x-\sqrt{x}\le0\)

\(\Rightarrow\) \(P\sqrt{x}\) đạt giá trị lớn nhất là \(0\) khi \(x=0\left(ktmđk\right)\)