Bài 1:
a: Thay x=36 vào A, ta được:
\(A=\dfrac{36-3\cdot6+16}{6-3}=\dfrac{36-18+16}{3}=\dfrac{34}{3}\)
b: \(B=\dfrac{2x-4\sqrt{x}+6-x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{x-5\sqrt{x}+6}{\left(\sqrt{x}-2\right)\cdot\sqrt{x}}=\dfrac{\sqrt{x}-3}{\sqrt{x}}\)