a: Thay x=25 vào E, ta được:
\(E=\dfrac{25+7}{\sqrt{25}-3}=\dfrac{32}{5-3}=\dfrac{32}{2}=16\)
b: \(F=\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{3}{3-\sqrt{x}}-\dfrac{x+3}{x-2\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{3}{\sqrt{x}-3}-\dfrac{x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-3\left(\sqrt{x}+1\right)-x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-3\sqrt{x}-3\sqrt{x}-3-x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-6\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}=\dfrac{-6}{\sqrt{x}-3}\)
c: \(A=E+F=\dfrac{-6}{\sqrt{x}-3}+\dfrac{x+7}{\sqrt{x}-3}=\dfrac{x+1}{\sqrt{x}-3}\)
|A|>A
=>A<0
=>\(\dfrac{x+1}{\sqrt{x}-3}< 0\)
=>\(\sqrt{x}-3< 0\)
=>\(\sqrt{x}< 3\)
=>0<=x<9
mà x là số nguyên lớn nhất
nên x=8
