a: ĐKXĐ: x>=0; x<>4
b: \(B=\dfrac{x-3\sqrt{x}+2+3x+6\sqrt{x}-2+5\sqrt{x}}{x-4}\)
\(=\dfrac{4x+8\sqrt{x}}{x-4}=\dfrac{4\sqrt{x}}{\sqrt{x}-2}\)
c: Khi x=25 thì \(B=\dfrac{4\cdot3}{5-2}=\dfrac{12}{3}=4\)
a/ ĐKXĐ : \(x>0,x\ne4\)
b/ \(B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+3\sqrt{x}\left(\sqrt{x}+2\right)-2+5\sqrt{x}}{x-4}\)
\(=\dfrac{x-2\sqrt{x}-\sqrt{x}+2+3x+6\sqrt{x}-2+5\sqrt{x}}{x-4}\)
\(=\dfrac{4x+8\sqrt{x}}{x-4}=\dfrac{4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{4\sqrt{x}}{\sqrt{x}-2}\)(đpcm)
c/ Thay x = 25 ( TMĐK ) vào B ta đc :
\(\dfrac{4.\sqrt{25}}{\sqrt{25}-2}=\dfrac{4.5}{5-2}=\dfrac{20}{3}\)
d/ \(B=12\Leftrightarrow\dfrac{4\sqrt{x}}{\sqrt{x}-2}=12\Rightarrow12\left(\sqrt{x}-2\right)=4\sqrt{x}\Leftrightarrow12\sqrt{x}-24-4\sqrt{x}=0\Leftrightarrow8\sqrt{x}=24\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\)
