a , \(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
\(P=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}+\dfrac{3\left(\sqrt{x}-1\right)}{x-1}-\dfrac{6\sqrt{x}-4}{x-1}\)
\(P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{x-1}\)
\(P=\dfrac{x-2\sqrt{x}+1}{x-1}\)
\(P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b , Với x = 9 ta có :
\(P=\dfrac{\sqrt{9}-1}{\sqrt{9}+1}=\dfrac{3-1}{3+1}=\dfrac{2}{4}=\dfrac{1}{2}\)
ĐKXĐ : \(x\ge0,x\ne1\)
Câu a : \(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-\left(6\sqrt{x}-4\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
Câu b : Thay \(x=9\) vào biểu thức P ta được :
\(P=\dfrac{\sqrt{9}-1}{\sqrt{9}+1}=\dfrac{3-1}{3+1}=\dfrac{2}{4}=\dfrac{1}{2}\)
Câu c : \(P< \dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}< \dfrac{1}{2}\)
\(\Leftrightarrow2\sqrt{x}-2< \sqrt{x}+1\)
\(\Leftrightarrow\sqrt{x}< 3\)
\(\Leftrightarrow x< 9\)
a) \(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}+\dfrac{3\left(\sqrt{x}-1\right)}{x-1}-\dfrac{6\sqrt{x}-4}{x-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-\left(6\sqrt{x}-4\right)}{x-1}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{x-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{x-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) Tại x = 9 ta có:
\(\dfrac{\sqrt{9}-1}{\sqrt{9}+1}=\dfrac{3-1}{3+1}=\dfrac{2}{4}=\dfrac{1}{2}\)
c) Để \(P< \dfrac{1}{2}\) thì
\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}< \dfrac{1}{2}\)
\(\Leftrightarrow2\left(\sqrt{x}-1\right)< \sqrt{x}+1\)
\(\Leftrightarrow2\sqrt{x}-2-\sqrt{x}-1< 0\)
\(\Leftrightarrow\sqrt{x}-3< 0\)
\(\Leftrightarrow\sqrt{x}< 3\)
\(\Leftrightarrow x< 9\)
Vậy x <9
