Question

Answer 1

a: \(B=\dfrac{x+\sqrt{x}-2-x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{-1}{\sqrt{x}-2}\)

Answer 2

a, \(B=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{2-\sqrt{x}}\)

\(\Rightarrow B=\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow B=\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow B=\dfrac{x+\sqrt{x}-2-x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow B=\dfrac{-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow B=\dfrac{-\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow B=\dfrac{-1}{\sqrt{x}-2}\)

\(\Rightarrow B=\dfrac{1}{2-\sqrt{x}}\)

b, Để B<0 thì \(\dfrac{1}{2-\sqrt{x}}< 0\Rightarrow2-\sqrt{x}< 0\Rightarrow\sqrt{x}>2\Rightarrow x>4\)

Answer 3