A ∪ B = (-∞; +∞)

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Giá bán của a sản phẩm là: 190a(nghìn đồng)

Lợi nhuận là: \(A=190a-\left(a^2+30a+4800\right)\)

\(=190a-a^2-30a-4800=-a^2+160a-4800\)

\(=-a^2+2\cdot a\cdot80-6400+1600\)

\(=-\left(a-80\right)^2+1600\)

Để không bị lỗ thì A>=0

=>\(1600-\left(a-80\right)^2\ge0\)

=>\(\left(a-80\right)^2\le1600\)

=>-40<=a-80<=a+80

=>40<=a<=120

=>Số sản phẩm nhiều nhất có thể được sản xuất để không bị lỗ là 120 sản phẩm

1) \(\overrightarrow{AM}=\overrightarrow{AD}+\overrightarrow{DM}\)

             \(=\overrightarrow{AD}+\dfrac{2}{3}\overrightarrow{DC}\)

             \(=\overrightarrow{AD}+\dfrac{2}{3}\left(\overrightarrow{AC}-\overrightarrow{AD}\right)\)

             \(=\dfrac{2}{3}\overrightarrow{AC}+\dfrac{1}{3}\overrightarrow{AD}\) (đpcm)

2) \(AC=BD=\sqrt{AB^2+AD^2}=\sqrt{4^2+2^2}=2\sqrt{5}\)

\(\overrightarrow{AC}.\overrightarrow{AD}=\dfrac{AC^2+AD^2-CD^2}{2}\)

               \(=\dfrac{20+4-16}{2}=4\)

3) Gọi O là tâm hình chữ nhật

\(\Rightarrow2\overrightarrow{OA}+\overrightarrow{OB}+2\overrightarrow{OC}+\overrightarrow{OD}=\overrightarrow{0}\)

Ta có:

\(2PA^2+PB^2+2PC^2+PD^2\)

\(=2\left(\overrightarrow{PO}+\overrightarrow{OA}\right)^2+\left(\overrightarrow{PO}+\overrightarrow{OB}\right)^2+2\left(\overrightarrow{PO}+\overrightarrow{OC}\right)^2+\left(\overrightarrow{PO}+\overrightarrow{OD}\right)^2\)

\(=6PO^2+2OA^2+OB^2+2OC^2+OD^2+2\overrightarrow{PO}\left(2\overrightarrow{OA}+\overrightarrow{OB}+2\overrightarrow{OC}+\overrightarrow{OD}\right)\)

\(=\)\(6PO^2+2OA^2+OB^2+2OC^2+OD^2\)

\(=6PO^2+6OA^2\left[OB=OD=OA=OC\right]\)

\(=6PO^2+6\left(\sqrt{5}\right)^2\)

\(=6PO^2+30\ge30\) 

Dấu "=" xảy ra \(\Leftrightarrow O\equiv P\) 

\(\Rightarrow\dfrac{1}{2PA^2+PB^2+2PC^2+PD^2}\le\dfrac{1}{30}\)

\(Max\dfrac{1}{2PA^2+PB^2+2PC^2+PD^2}=\dfrac{1}{30}\Leftrightarrow P\equiv O\)

1) \(\sqrt{2x-1}=\sqrt{x^2+4x-4}\left(Đk:x\ge\dfrac{1}{2}\right)\)

\(\Leftrightarrow2x-1=x^2+4x-4\)

\(\Leftrightarrow x^2+2x-3=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(TM\right)\\x=-3\left(L\right)\end{matrix}\right.\)

Vậy \(S=\left\{1\right\}\)

2) \(\sqrt{x^2-4x+3}=x-3\left(Đk:x\ge3\right)\)

\(\Leftrightarrow\left(x-3\right)^2=x^2-4x+3\)

\(\Leftrightarrow x^2-6x+9=x^2-4x+3\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

Vậy \(S=\left\{3\right\}\)

Để phương trình: \(x^2-2\left(m-1\right)x+4m+8=0\) có nghiệm

\(\Rightarrow\Delta\ge0\)

\(\Leftrightarrow4\left(m-1\right)^2-4\left(4m+8\right)\ge0\)

\(\Leftrightarrow4m^2-8m+4-16m-32\ge0\)

\(\Leftrightarrow4m^2-24m-28\ge0\)

\(\Leftrightarrow m^2-6m-7\ge0\)

\(\Leftrightarrow\left(m+1\right)\left(m-7\right)\ge0\)

\(\Rightarrow m\in(-\infty;-1]\cup[7;+\infty)\)

Tọa độ đỉnh là:

\(\begin{cases}x=-\frac{b}{2a}=\frac{-\left(-4\right)}{2\cdot1}=\frac42=2\\ y=-\frac{b^2-4ac}{4a}=-\frac{\left(-4\right)^2-4\cdot1\cdot3}{4\cdot1}=-\frac{16-12}{4}=-1\end{cases}\)

=>Trục đối xứng là x=2

Vẽ đồ thị là:

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