a: Khi x=9 thì \(A=\dfrac{3-2}{3-1}=\dfrac{1}{2}\)
b: \(B=\dfrac{x-\sqrt{x}-5\sqrt{x}-3+4\sqrt{x}+4}{x-1}=\dfrac{x-2\sqrt{x}+1}{x-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
a) Khi x = 9 TMĐK thì: \(A=\dfrac{\sqrt{9}-2}{\sqrt{9}-2}=\dfrac{3-2}{3-1}=\dfrac{1}{2}\)
Vậy \(x=9\) thì \(A=\dfrac{1}{2}\).
b) Với \(x\ge0 ; x\ne 1\) thì:
\(B=\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{5\sqrt{x}+3}{1-x}+\dfrac{4}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{5\sqrt{x}+3}{x-1}+\dfrac{4}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-5\sqrt{x}-3+4\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}-5\sqrt{x}-3+4\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\) (đpcm)
