ĐKXĐ: \(x\ge0\)
\(x-2\sqrt{x}+5\)
\(\)\(=x-2\sqrt{x}+1+4\)
\(=\left(\sqrt{x}\right)^2-2\sqrt{x}+1+4\)
\(=\left(\sqrt{x}-1\right)^2+4\)
Ta có : \(\left(\sqrt{x}-1\right)^2\ge0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+4\ge4\)
\(\Leftrightarrow\frac{1}{\left(\sqrt{x}-1\right)^2+4}\le\frac{1}{4}\)
\(\Leftrightarrow\sqrt{\frac{1}{\left(\sqrt{x}-1\right)^2+4}}\le\frac{1}{2}\)
Vậy maxP\(=\frac{1}{2}\)
Và dấu"=" xảy ra khi \(x=1\)
mình dặt P là biểu thức như trên
(xin lỗi nhiều nha)
Đặt \(A=\sqrt{\frac{1}{x-2\sqrt{x}+5}}=\sqrt{\frac{1}{\left(x-2\sqrt{x}+1\right)+4}}=\sqrt{\frac{1}{\left(\sqrt{x}-1\right)^2+4}}\)
Ta có: \(\left(\sqrt{x}-1\right)^2\ge0\)
\(\Rightarrow\left(\sqrt{x}-1\right)^2+4\ge4\)
\(\Rightarrow\frac{1}{\left(\sqrt{x}-1\right)^2+4}\le\frac{1}{4}\)
\(\Rightarrow\sqrt{\frac{1}{\left(\sqrt{x}-1\right)^2+4}}\le\frac{1}{2}\)
hay \(A\le0,5\)
Vậy: Max A = 0,5 khi \(x=1\)
