Question

Answer 1

a, x >0 ;x ≠ 1

\(P=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{x-1}\\ =\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{x-1}\\ =\dfrac{x-2\sqrt{x}+1}{x-1}\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{x-1}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\ b,thayx=81\left(thoamanđk\right)\\ \dfrac{\sqrt{81}-1}{\sqrt{81}+1}\\ =\dfrac{9-1}{9+1}=\dfrac{8}{10}=\dfrac{4}{5}\\ c,P=\dfrac{1}{2}\\ \dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{1}{2}\\ 2\sqrt{x}-2=\sqrt{x}+1\\ 2\sqrt{x}-\sqrt{x}=2+1\\ \sqrt{x}=3\\ x=9\left(thoamanđk\right)\)