Question

[Phong Lê]

  …….

Answer 1

[Nguyễn Lê Phước Thịnh]

1: Thay x=36 vào A, ta được:

\(A=\dfrac{36-3\cdot4+16}{4-3}=\dfrac{52-12}{1}=40\)

2: \(B=\dfrac{2x-4\sqrt{x}+6}{x-2\sqrt{x}}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)

\(=\dfrac{2x-4\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)

\(=\dfrac{2x-4\sqrt{x}+6-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\cdot\sqrt{x}}\)

\(=\dfrac{2x-4\sqrt{x}+6-x-\sqrt{x}}{\left(\sqrt{x}-2\right)\cdot\sqrt{x}}\)

\(=\dfrac{x-5\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}}\)

3: \(P=A\cdot B=\dfrac{\sqrt{x}-3}{\sqrt{x}}\cdot\dfrac{x-3\sqrt{x}+16}{\sqrt{x}-3}\)

\(=\dfrac{x-3\sqrt{x}+16}{\sqrt{x}}\)

\(=\sqrt{x}-3+\dfrac{16}{\sqrt{x}}\)

\(=\sqrt{x}+\dfrac{16}{\sqrt{x}}-3>=2\cdot\sqrt{\sqrt{x}\cdot\dfrac{16}{\sqrt{x}}}-3=5\)

Dấu '=' xảy ra khi \(\sqrt{x}^2=16\)

=>x=16