B)
\(B=\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\left(x\ge0,x\ne25\right)\)
\(=\left[\dfrac{15-\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}+\dfrac{2\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right]:\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)
\(=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}-5\right)\left(\sqrt{x}++5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+7}\)
\(=\dfrac{1}{\sqrt{x}+1}\)
c)
\(p=A.B=\dfrac{4\left(\sqrt{x}+1\right)}{25-x}.\dfrac{1}{\sqrt{x}+1}\)
để p nguyên lớn nhất
⇒p là ước dương nhỏ nhất của 4
⇒25-x =1⇒25-1=x⇒x=24
vậy p nguyên lớn nhất \(=\dfrac{4}{25-24}=4\)
\(khix=24\)
a)
\(A=\dfrac{4\left(\sqrt{x}+1\right)}{25-x}khix=9\)
\(\Rightarrow A=\dfrac{4\left(\sqrt{9}+1\right)}{25-9}=\dfrac{4\left(3+1\right)}{16}=1\)
