Question

Answer 1

a)Ta có\(x-\sqrt{x}-6=0\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\left(nhận\right)\\\sqrt{x}=-2\left(loại\right)\end{matrix}\right.\Leftrightarrow x=9}\)

\(A=\dfrac{x-\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{9-3+2}{6}=\dfrac{8}{6}=\dfrac{4}{3}\)

b)\(B=\dfrac{x-4+2\sqrt{x}+6-3\sqrt{x}-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)