ĐK: x\(\ge0,x\ne1\)
a) \(A=\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}+1}=\dfrac{x\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(x-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{x\sqrt{x}-x-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x\sqrt{x}+1-x\sqrt{x}+x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
b) Thay x=\(\dfrac{1}{4}\) vào A thì A=\(\dfrac{\sqrt{\dfrac{1}{4}}}{\sqrt{\dfrac{1}{4}}-1}=\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}-1}=\dfrac{\dfrac{1}{2}}{-\dfrac{1}{2}}=-1\)
Vậy khi x=\(\dfrac{1}{4}\) thì A=-1
