a) \(đk:x\ne1\)
b) \(đk:x\ne1\)
thay x=9(tm) vào A ta có: A=\(\frac{\sqrt{9}-1}{\sqrt{9}+1}=\frac{3-1}{3+1}=\frac{2}{4}=\frac{1}{2}\)
c) P=A.B
P=\(\frac{\sqrt{x}-1}{\sqrt{x}+1}.\left(\frac{\sqrt{x}+3}{\sqrt{x}+1}-\frac{4}{1-\sqrt{x}}+\frac{5-x}{x-1}\right)\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}+1}.\left(\frac{\sqrt{x}+3}{\sqrt{x}+1}+\frac{4}{\sqrt{x}-1}+\frac{5-x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}+1}.\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)+4\left(\sqrt{x}+1\right)+5-x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}+1}.\frac{x+2\sqrt{x}-3+4\sqrt{x}+4+5-x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{6\sqrt{x}+6}{\left(\sqrt{x}+1\right)^2}\)
\(=\frac{6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2}\)
\(=\frac{6}{\sqrt{x}+1}\)