\(2x+\left|2x-5\right|=2x+\left|5-2x\right|\ge2x+5-2x=5.\Rightarrow A_{min}=5.\text{Dâu "=" xay }ra\Leftrightarrow2x-5\ge0\Leftrightarrow x\le2,5\)
\(M=\left|x\right|+\left|x-1\right|=\left|x\right|+\left|1-x\right|\ge x+1-x=1\Rightarrow M_{min}=1.\text{Dâu "=" xay ra}\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\1-x\ge0\end{matrix}\right.\Leftrightarrow0\le x\le1\)
\(A=x-\sqrt{x}\Leftrightarrow A+\frac{1}{4}=x-\sqrt{x}+\frac{1}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\Rightarrow A+\frac{1}{4}\ge0\Rightarrow A_{min}=\frac{-1}{4}.\text{Dâus "=" xay ra khi:}x=\frac{1}{4}\)
Bài 1:
Sửa đề :v
\(B=x\left(x-3\right)\left(x-1\right)\left(x-4\right)\)
\(B=\left(x^2-4x\right)\left(x^2-4x+3\right)\)
Đặt \(x^2-4x=t\)
\(B=t\left(t+3\right)\)
\(B=t^2+3t=t^2+2\cdot t\cdot\frac{3}{2}+\frac{9}{4}-\frac{9}{4}=\left(t+\frac{3}{2}\right)^2-\frac{9}{4}\ge\frac{-9}{4}\forall t\)
Dấu "=" xảy ra \(\Leftrightarrow t=\frac{-3}{2}\Leftrightarrow x^2-4x=\frac{-3}{2}\Leftrightarrow x=\frac{4\pm\sqrt{10}}{2}\)
Bài 2: Mình nghĩ nên sửa đề tìm min \(A=\left|2x\right|+\left|2x-5\right|\)
Bài 3:
\(M=\left|x\right|+\left|x-1\right|\)
\(M=\left|x\right|+\left|1-x\right|\ge\left|x+1-x\right|=1\)
Dấu "=" xảy ra \(\Leftrightarrow x\left(1-x\right)\ge0\Leftrightarrow0\le x\le1\)
Bài 4:
\(A=x-\sqrt{x}\)
Do điều kiện \(x\ge0\)
\(\Rightarrow A\ge0+0=0\)
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
