Question

Bài 1:

  

Answer 1

Bài I:

1: Thay x=4 vào B, ta được:

\(B=\dfrac{1}{\sqrt{4}-3}=\dfrac{1}{2-3}=-1\)

2: M=A:B

\(=\left(\dfrac{x+\sqrt{x}+10}{x-9}-\dfrac{1}{\sqrt{x}-3}\right):\dfrac{1}{\sqrt{x}-3}\)

\(=\left(\dfrac{x+\sqrt{x}+10}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{1}{\sqrt{x}-3}\right)\cdot\dfrac{\sqrt{x}-3}{1}\)

\(=\dfrac{x+\sqrt{x}+10-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)}{1}\)

\(=\dfrac{x+7}{\sqrt{x}+3}\)

3: \(M=\dfrac{x+7}{\sqrt{x}+3}=\dfrac{x-9+16}{\sqrt{x}+3}\)

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

\(=\sqrt{x}+3+\dfrac{16}{\sqrt{x}+3}-6\)

=>\(M>=2\cdot\sqrt{\left(\sqrt{x}+3\right)\cdot\dfrac{16}{\sqrt{x}+3}}-6=2\cdot4-6=2\)

Dấu '=' xảy ra khi \(\sqrt{x}+3=\sqrt{16}=4\)

=>x=1