Theo mk nghĩ thì đề bài fải như thế này:

\(\left(4x^5+2x^4+4x^3-x^2-1\right):\left(2x^3+x-1\right)\)

Kết quả của phép chia trên là: \(2x^2+x+1\)

Ta có: \(2x^2+x+1=2\left(x^2+\frac{1}{2}x+\frac{1}{2}\right)\)

\(=2\left(x^2+\frac{1}{2}x+\frac{1}{16}+\frac{7}{16}\right)\)

\(=2\left(x+\frac{1}{4}\right)^2+\frac{7}{8}\ge\frac{7}{8}\forall x\)

=> Min = 7/8 tại \(2\left(x+\frac{1}{4}\right)^2=0\Rightarrow x=-\frac{1}{4}\)

=.= hok tốt!!

c: \(-x^2+2x-2=-\left(x-1\right)^2-1\le-1\forall x\)

\(\Leftrightarrow V\ge-1\forall x\)

Dấu '=' xảy ra khi x=1

\(\dfrac{4x^5+2x^4+4x^3-x-1}{2x^3+x-1}\)

\(=\dfrac{4x^5+2x^3-2x^2+2x^4+x^2-x+2x^3+x-1+x^2}{2x^3+x-1}\)

\(=2x^2+x+1+\dfrac{x^2}{2x^3+x-1}\)

=>Thương là A=2x^2+x+1

=2(x^2+1/2x+1/2)

=2(x^2+2*x*1/4+1/16+7/16)

=2(x+1/4)^2+7/8>=7/8

Dấu = xảy ra khi x=-1/4

\(A=\left(x-1\right)^2+8\ge8\\ A_{min}=8\Leftrightarrow x=1\\ B=\left(x+3\right)^2-12\ge-12\\ B_{min}=-12\Leftrightarrow x=-3\\ C=x^2-4x+3+9=\left(x-2\right)^2+8\ge8\\ C_{min}=8\Leftrightarrow x=2\\ E=-\left(x+2\right)^2+11\le11\\ E_{max}=11\Leftrightarrow x=-2\\ F=9-4x^2\le9\\ F_{max}=9\Leftrightarrow x=0\)

Giups mik vs

...Thương: \(A=2x^2+x+1=\left(\sqrt{2}\cdot x+\dfrac{\sqrt{2}}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}\)

''='' xảy ra <=> \(x=-\dfrac{1}{4}\)

Vậy GTNN của thương trong phép chia này là \(\dfrac{7}{8}\)

P/s: Nếu chỗ lm phép chia k nhìn thấy thì tự lm ra nháp nhá :v

a, \(A=x^4-2x^3+2x^2-2x+3\)

\(=\left(x^4+2x^2+1\right)-\left(2x^3+2x\right)+2\)

\(=\left(x^2+1\right)^2-2x\left(x^2+1\right)+2\)

\(=\left(x^2+1\right)\left(x^2-2x+1\right)+2\)

\(=\left(x^2+1\right)\left(x-1\right)^2+2\)

Vì \(\hept{\begin{cases}x^2\ge0\\\left(x-1\right)^2\ge0\end{cases}\Rightarrow\hept{\begin{cases}x^2+1\ge1\\\left(x-1\right)^2\ge0\end{cases}\Rightarrow}\left(x^2+1\right)\left(x-1\right)^2\ge0}\)

\(\Rightarrow A=\left(x^2+1\right)\left(x-1\right)^2+2\ge2\)

Dấu "=" xảy ra khi x = 1

Vậy Amin = 2 khi x = 1

b, \(B=4x^2-2\left|2x-1\right|-4x+5=\left(4x^2-4x+1\right)-2\left|2x-1\right|+4=\left(2x-1\right)^2-2\left|2x-1\right|+4\)

đề sai ko

c, \(C=4-x^2+2x=-\left(x^2-2x+1\right)+5=-\left(x-1\right)^2+5\)

Vì \(-\left(x-1\right)^2\le0\Rightarrow C=-\left(x-1\right)^2+5\le5\)

Dấu "=" xảy ra khi x=1

Vậy Cmin = 5 khi x = 1

2/

+) \(D=-x^2-y^2+x+y+3=-\left(x^2-x+\frac{1}{4}\right)-\left(y^2-y+\frac{1}{4}\right)+\frac{7}{2}=-\left(x-\frac{1}{2}\right)^2-\left(y-\frac{1}{2}\right)^2+\frac{7}{2}\)

Vì \(\hept{\begin{cases}-\left(x-\frac{1}{2}\right)^2\le0\\-\left(y-\frac{1}{2}\right)^2\le0\end{cases}\Rightarrow-\left(x-\frac{1}{2}\right)^2-\left(y-\frac{1}{2}\right)^2\le0}\Rightarrow D=-\left(x-\frac{1}{2}\right)^2-\left(y-\frac{1}{2}\right)^2+\frac{7}{2}\le\frac{7}{2}\)

Dấu "=" xảy ra khi x=y=1/2

Vậy Dmax=7/2 khi x=y=1/2

+) Đề sai

+)bài này là tìm min 

 \(G=x^2-3x+5=\left(x^2-3x+\frac{9}{4}\right)+\frac{11}{4}=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)

Dấu "=" xảy ra khi x=3/2

Vậy Gmin=11/4 khi x=3//2