Đặt \(A=\frac{7x-4\sqrt{x}}{\sqrt{x}+2}\)
\(=\frac{7x+14\sqrt{x}-18\sqrt{x}-36+36}{\sqrt{x}+2}\)
\(=7\sqrt{x}-18+\frac{36}{\sqrt{x}+2}=7\sqrt{x}+14+\frac{36}{\sqrt{x}+2}-32\)
\(=7\left(\sqrt{x}+2\right)+\frac{36}{\sqrt{x}+2}-32\)
Ta có: \(7\left(\sqrt{x}+2\right)+\frac{36}{\sqrt{x}+2}\ge2\cdot\sqrt{7\left(\sqrt{x}+2\right)\cdot\frac{36}{\sqrt{x}+2}}=2\cdot6\sqrt7=12\sqrt7\) ∀x thỏa mãn ĐKXĐ
=>\(7\left(\sqrt{x}+2\right)+\frac{36}{\sqrt{x}+2}-32\ge12\sqrt7-32\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi \(7\left(\sqrt{x}+2\right)^2=36\)
=>\(\left(\sqrt{x}+2\right)^2=\frac{36}{7}=\frac{252}{49}=\left(\frac{6\sqrt7}{7}\right)^2\)
=>\(\sqrt{x}+2=\frac{6\sqrt7}{7}\)
=>\(\sqrt{x}=\frac{6\sqrt7}{7}-2=\frac{6\sqrt7-14}{7}\)
=>\(x=\left(\frac{6\sqrt7-14}{7}\right)^2=\frac{36\cdot7-2\cdot6\sqrt7\cdot14+196}{49}=\frac{448-168\sqrt7}{49}=\frac{64-24\sqrt7}{7}\)
