\(a,A=\dfrac{\sqrt{x}+3}{x-4}\left(x\ge0;x\ne4\right)\)
Thay \(x=9\) vào \(A\), ta được:
\(A=\dfrac{\sqrt{9}+3}{9-4}=\dfrac{3+3}{5}=\dfrac{6}{5}\)
\(---\)
\(b,B=\dfrac{-4}{x-4}+\dfrac{1}{\sqrt{x}-2}\left(x\ge0;x\ne4\right)\)
\(=\dfrac{-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{-4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{1}{\sqrt{x}+2}\)
\(---\)
\(c,P=\dfrac{A}{B}\left(x>4\right)\)
\(=\dfrac{\sqrt{x}+3}{x-4}:\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}-2+5}{\sqrt{x}-2}\)
\(=1+\dfrac{5}{\sqrt{x}-2}\)
Khi đó: \(x>4\Leftrightarrow\sqrt{x}>2\)
\(\Leftrightarrow\sqrt{x}-2>0\)
\(\Leftrightarrow\dfrac{5}{\sqrt{x}-2}>0\)
\(\Leftrightarrow1+\dfrac{5}{\sqrt{x}-2}>1\)
hay \(P>1\) khi \(x>4\)
#\(Toru\)
