Bài 3:
a: Thay x=36 vào A, ta được:
\(A=\dfrac{36-6+4}{2\cdot6+1}=\dfrac{30+4}{13}=\dfrac{34}{13}\)
b: \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}+\dfrac{\sqrt{x}+2}{\sqrt{x}+3}+\dfrac{6\left(\sqrt{x}-1\right)}{9-x}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)-6\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+2\sqrt{x}-3+x-\sqrt{x}-6-6\sqrt{x}+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2x-5\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2\sqrt{x}+1}{\sqrt{x}+3}\)
\(P=A\cdot B=\dfrac{2\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{x-\sqrt{x}+4}{2\sqrt{x}+1}=\dfrac{x-\sqrt{x}+4}{\sqrt{x}+3}\)
c: \(P=\dfrac{x-\sqrt{x}+4}{\sqrt{x}+3}=\dfrac{x+3\sqrt{x}-4\sqrt{x}-12+16}{\sqrt{x}+3}\)
\(=\sqrt{x}-4+\dfrac{16}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{16}{\sqrt{x}+3}-7\)
=>\(P>=2\cdot\sqrt{\left(\sqrt{x}+3\right)\cdot\dfrac{16}{\sqrt{x}+3}}-7=2\cdot4-7=1\)
Dấu '=' xảy ra khi \(\sqrt{x}+3=\sqrt{16}=4\)
=>x=1
