1: Thay x=36 vào A, ta được:
\(A=\dfrac{6+2}{6}=\dfrac{8}{6}=\dfrac{4}{3}\)
2: \(B=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
3: P=A:B
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{x-4}{x}\)
\(x\cdot P< =10\sqrt{x}-29-\sqrt{x-25}\)
=>\(x-4< =10\sqrt{x}-29-\sqrt{x-25}\)
=>\(x+25-10\sqrt{x}+\sqrt{x-25}< =0\)
=>\(\left(\sqrt{x}-5\right)^2+\sqrt{x-25}< =0\)
=>\(\left\{{}\begin{matrix}\sqrt{x}-5=0\\\sqrt{x-25}=0\end{matrix}\right.\)
=>x=25(nhận)
