\(1,x=9\Rightarrow A=\dfrac{\sqrt{9}+3}{\sqrt{9}-4}=\dfrac{3+3}{3-4}=-6\)
\(2,B=\dfrac{\sqrt{x}+3}{\sqrt{x}+4}+\dfrac{5\sqrt{x}+12}{x-16}\left(dk:x\ge0,x\ne16\right)\\ =\dfrac{\sqrt{x}+3}{\sqrt{x}+4}+\dfrac{5\sqrt{x}+12}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}\\ =\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)+5\sqrt{x}+12}{x-16}\\ =\dfrac{x-\sqrt{x}-12+5\sqrt{x}+12}{x-16}\\ =\dfrac{x+4\sqrt{x}}{x-16}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}\\ =\dfrac{\sqrt{x}}{\sqrt{x}-4}\)
\(c,\dfrac{A}{B}=2\Leftrightarrow\dfrac{\sqrt{x}+3}{\sqrt{x}-4}:\dfrac{\sqrt{x}}{\sqrt{x}-4}=2\\ \Leftrightarrow\dfrac{\sqrt{x}+3}{\sqrt{x}}=2\Leftrightarrow\sqrt{x}+3-2\sqrt{x}=0\Leftrightarrow-\sqrt{x}=-3\Leftrightarrow x=9\left(tmdk\right)\)
Vậy \(x=9\) thì \(\dfrac{A}{B}=2\)
1) Thay x=9 vào A ta có:
\(A=\dfrac{\sqrt{9}+3}{\sqrt{9}-4}=\dfrac{3+3}{3-4}=\dfrac{6}{-1}=-6\)
2) \(B=\dfrac{\sqrt{x}+3}{\sqrt{x}+4}+\dfrac{5\sqrt{x}+12}{x-16}\)
\(B=\dfrac{\sqrt{x}+3}{\sqrt{x}+4}+\dfrac{5\sqrt{x}+12}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\)
\(B=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}+\dfrac{5\sqrt{x}+12}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\)
\(B=\dfrac{x+3\sqrt{x}-4\sqrt{x}-12+5\sqrt{x}+12}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\)
\(B=\dfrac{x+4\sqrt{x}}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\)
\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\)
\(B=\dfrac{\sqrt{x}}{\sqrt{x}-4}\)
3) Ta có: \(\dfrac{A}{B}=2\) khi:
\(\dfrac{\dfrac{\sqrt{x}+3}{\sqrt{x}-4}}{\dfrac{\sqrt{x}}{\sqrt{x}-4}}=2\)
\(\Leftrightarrow\dfrac{\sqrt{x}+3}{\sqrt{x}-4}\cdot\dfrac{\sqrt{x}-4}{\sqrt{x}}=2\)
\(\Leftrightarrow\dfrac{\sqrt{x}+3}{\sqrt{x}}=2\)
\(\Leftrightarrow\sqrt{x}+3=2\sqrt{x}\)
\(\Leftrightarrow2\sqrt{x}-\sqrt{x}=3\)
\(\Leftrightarrow\sqrt{x}=3\)
\(\Leftrightarrow x=9\left(tm\right)\)
