Ta có:\(B=x-4\sqrt{x}+10x-4x+1\)
\(=7x-4\sqrt{x}+1\)
\(=7\left(x-\frac{4}{7}\sqrt{x}+1\right)\)
\(=7\left(x-2.\frac{2}{7}\sqrt{x}+\frac{4}{49}-\frac{4}{49}+1\right)\)
\(=7\left[\left(\sqrt{x}-\frac{2}{7}\right)^2+\frac{45}{49}\right]\)
\(=7\left(\sqrt{x}-\frac{2}{7}\right)^2+\frac{45}{7}\)
Lại có:\(\left(\sqrt{x}-\frac{2}{7}\right)^2\ge0,\forall x\ge0\)
\(\Leftrightarrow7\left(\sqrt{x}-\frac{2}{7}\right)^2\ge0\)
\(\Leftrightarrow7\left(\sqrt{x}-\frac{2}{7}\right)^2+\frac{45}{7}\ge\frac{45}{7}\)
\(\Rightarrow min_B=\frac{45}{7}\) khi \(\sqrt{x}-\frac{2}{7}=0\Leftrightarrow\sqrt{x}=\frac{2}{7}\Leftrightarrow x=\frac{4}{49}\)
Đề bạn sửa lại: \(B=x-4\sqrt{x}+10\)
Ta có: \(B=x-2\sqrt{x}.2+4+6=\left(\sqrt{x}-2\right)^2+6\)
Lại có: \(\left(\sqrt{x}-2\right)^2\ge0,\forall x\ge0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+6\ge6\)
\(\Rightarrow Min_B=6\) khi \(\sqrt{x}-2=0\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)
