a: \(A=\sqrt{x}-\dfrac{x+2}{\sqrt{x}+1}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)-x-2}{\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}-x-2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
Thay x=9 vào A, ta được:
\(A=\dfrac{3-2}{3+1}=\dfrac{1}{4}\)
b: P=A:B
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}:\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}-4}{1-x}\right)\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}:\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}:\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x-4}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
c: P=1/2
=>\(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=\dfrac{1}{2}\)
=>\(2\left(\sqrt{x}-1\right)=\sqrt{x}+2\)
=>\(2\sqrt{x}-2=\sqrt{x}+2\)
=>\(\sqrt{x}=4\)
=>x=16(nhận)
