\(y=x-4\sqrt{x}-1.\)
\(y=x-2\cdot2\sqrt{x}+2-2-1\)
\(y=\left(\sqrt{x}-2\right)^2-2-1\)
\(y=\left(\sqrt{x}-2\right)^2-3\)
có \(\left(\sqrt{x}-2\right)^2\ge0\)
\(\Rightarrow\left(\sqrt{x}-2\right)^2-3\ge-3\)
\(\Rightarrow GTNNy=-3\)
với \(\left(\sqrt{x}-2\right)^2=0;x=4\)
\(y=x-4\sqrt{x}-1\)
\(y=\left(\sqrt{x}\right)^2-2.\sqrt{x}.2+4-5\)
\(y=\left(\sqrt{x}-2\right)^2-5>5\)HOẶC=5
\(=>y_{min_{ }}=5< =>\sqrt{x}=2=>x=4\)
ý ý :)) tuoi là sai rồi :) làm theo bạn kia đi ạ :))) xin lỗi vì sự sai sót ngu ng' của mik
\(y=x-4\sqrt{x}-1\)
\(=\left(\sqrt{x}\right)^2-2.\sqrt{x}.2+4-5\)
\(=\left(\sqrt{x}-2\right)^2-5\ge-5\)
(Dấu "="\(\Leftrightarrow\sqrt{x}-2=0\Leftrightarrow x=4\))
Vậy \(y_{min}=-5\Leftrightarrow x=4\)