Question

Cho P = \([\dfrac{\sqrt{x}}{\sqrt{x}-2}\)- \(\dfrac{4}{x-2\sqrt{x}}]\) \([\dfrac{1}{\sqrt{x}+2}\)+ \(\dfrac{4}{x-4}]\)

với x \(\ge\) 0, x \(\ne\) 4

a) Rút gọn P

b) Khi x = 4 + 2\(\sqrt{3}\). Tính P

c) Tìm x để P dương

( Giải hộ mình nha, mình đang cần gấp)

Answer 1

a) điều kiện : \(x>0;x\ne4\)

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

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

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

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

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

\(x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}+1\right)^2}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\) \(\left(x>0\right)\)

thay vào P ta có \(P=\dfrac{\sqrt{3}+1+2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}+1-2\right)}=\dfrac{\sqrt{3}+3}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}+3}{2}\)

\(P>0\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\)

ta có : \(\sqrt{x}+2>0\) và \(\sqrt{x}>0\) \(\left(x>0\right)\)

\(\Rightarrow p>0\) thì \(\sqrt{x}-2>0\Leftrightarrow\sqrt{x}>2\Leftrightarrow x>4\)

vậy \(x>4\) thì P > 0