Question

1, Cho biết:

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

a, Rút gọn P

b, tìm x để P< 0

c, tính p khi x = 4 -\(2\sqrt{3}\)

Answer 1

\(ĐKXĐ:x>0;x\ne1\)

\(a.P=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}}.\dfrac{1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(b.P< 0\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}< 0\)

Do : \(x>0;x\ne1\Rightarrow\left\{{}\begin{matrix}\sqrt{x}+1>0\\\sqrt{x}>0\end{matrix}\right.\)

\(\Rightarrow\sqrt{x}-1< 0\)

\(\Leftrightarrow x< 1\)

Kết hợp ĐKXĐ : \(0< x< 1\)

\(c.x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}-1\)

Thay : \(x=4-2\sqrt{3}\left(TMĐKXĐ\right)\)vào bt P , ta có :

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