\(B=2x^2+8x+1\)

\(=2\times\left(x^2+2\times x\times2+2^2-2^2+\frac{1}{2}\right)\)

\(=2\times\left[\left(x+2\right)^2-\frac{7}{2}\right]\)

\(\left(x+2\right)^2\ge0\)

\(\left(x+2\right)^2-\frac{7}{2}\ge-\frac{7}{2}\)

\(2\times\left[\left(x+2\right)^2-\frac{7}{2}\right]\ge-7\)

Vậy Min B = -7 khi x = -2

\(C=\dfrac{x^2+2-\left(x^2-2x+1\right)}{x^2+2}=1-\dfrac{\left(x-1\right)^2}{x^2+2}\le1\)

\(C_{max}=1\) khi \(x=1\)

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

\(C_{min}=-\dfrac{1}{2}\) khi \(x=-2\)

Nhập Mode 7 , chạy trong khoản trung lập (-10;10)

tìm đc \(\begin{cases} C max = 1 khi x=1\\C min =-\dfrac{1}{2} khi x=-2 \end{cases}\)

Dùng cách này bạn giải trắc nghiệm sẽ nhanh hơn 

a) \(A=2\left|x-3\right|+\left|2x-10\right|=\left|2x-3\right|+\left|10-2x\right|\ge\left|2x-3+10-2x\right|=7\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(2x-3\right)\left(10-2x\right)\ge0\)\(\Leftrightarrow\)\(\frac{3}{2}\le x\le5\)

b) \(B\left|\frac{1}{4}x-8\right|+\left|2-\frac{1}{4}x\right|\ge\left|\frac{1}{4}x-8+2-\frac{1}{4}x\right|=6\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(\frac{1}{4}x-8\right)\left(2-\frac{1}{4}x\right)\ge0\)\(\Leftrightarrow\)\(8\le x\le32\)

\(A=2x^2-3x+2=2\left(x^2-\frac{3}{2}x\right)+2\)

\(=2\left(x^2-2.\frac{3}{4}x+\frac{9}{16}-\frac{9}{16}\right)+2=2\left(x-\frac{3}{4}\right)^2-\frac{9}{8}+2\ge\frac{7}{8}\)

Dấu ''='' xảy ra khi x = 3/4 

Vậy GTNN của A bằng 7/8 tại x = 3/4 

Ta có: A = 2x² + 4x + 5 = 2(x² + 2x + 1) + 3 = 2(x + 1)² + 3 \(\ge\)3 \(\forall\)x

Dấu "=" xảy ra <=> x + 1 = 0 <=> x = -1

Vậy MinA = 3 <=> x = -1

\(2x^2+4x+5\)

\(=2\left(x^2+2x+\frac{5}{2}\right)\)

\(=2\left(x^2+2x+1+\frac{3}{2}\right)\)

\(=2\left[\left(x+1\right)^2+\frac{3}{2}\right]\)

\(=2\left(x+1\right)^2+3\ge3\)

Dấu '' = '' xảy ra khi 

\(\Leftrightarrow2\left(x+1\right)^2=0\)

\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Vậy............................

P/s : sai thì thôi nha