\(A=\frac{x+7}{\sqrt{x}-3}\)
\(\Leftrightarrow A\left(\sqrt{x}-3\right)=x+7\)
\(\Leftrightarrow x-\sqrt{x}\cdot A+3A+7=0\)
Để phương trình có nghiệm thì:
\(\Delta=A^2-4\left(3A+7\right)\ge0\)
\(\Leftrightarrow A^2-12A-28\ge0\)
\(\Leftrightarrow\left(A-6\right)^2-64\ge0\)
\(\Leftrightarrow\left(A-6\right)^2\ge64\)
\(\Leftrightarrow A-6\ge8\)
\(\Leftrightarrow A\ge14\)
Dấu "=" xảy ra \(\Leftrightarrow x=49\)
\(A=\frac{x+7}{\sqrt{x}-3}\) \(\left(x>9\right)\)
Ta có: \(A=\frac{x+7}{\sqrt{x}-3}=\frac{x-9+16}{\sqrt{x}-3}=\frac{x^2-3^2}{\sqrt{x}-3}+\frac{16}{\sqrt{x}-3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}-3}+\frac{16}{\sqrt{x}-3}=\sqrt{x}+3+\frac{16}{\sqrt{x}-3}\)
\(=\sqrt{x}-3+\frac{16}{\sqrt{x}-3}+6\)
Vì \(x>9\) nên \(\sqrt{x}-3>0\) và \(\frac{16}{\sqrt{x}-3}>0\)
Áp dụng BĐT AM-GM vào 2 số trên, ta có:
\(\sqrt{x}-3+\frac{16}{\sqrt{x}-3}\ge2\sqrt{\left(\sqrt{x}-3\right)\cdot\frac{16}{\sqrt{x}-3}}=2\sqrt{16}=2\cdot4=8\) \(\forall x>9\)
\(\Rightarrow\sqrt{x}-3+\frac{16}{\sqrt{x}-3}+6\ge8+6\) \(\forall x>9\)
\(\Rightarrow A\ge14\) \(\forall x>9\)
Dấu bằng xảy ra \(\Leftrightarrow\sqrt{x}-3=\frac{16}{\sqrt{x}-3}\Rightarrow\left(\sqrt{x}-3\right)^2=16\Leftrightarrow\sqrt{x}-3=4\) (vì \(\sqrt{x}-3>0\))
\(\Leftrightarrow\sqrt{x}=7\Leftrightarrow x=49\) (tmđk)
Vậy \(A_{min}=14\) khi \(x=49\).
