Đặt \(A=\left|2x-3\right|+2\left|x-1\right|\)

\(\Rightarrow A=\left|2x-3\right|+\left|2x-2\right|=\left|2x-3\right|+\left|2-2x\right|\)

\(\Rightarrow A\ge\left|2x-3+2-2x\right|=\left|-1\right|=1\)

Dấu " = " xảy ra \(\Leftrightarrow\left(2x-3\right)\left(2-2x\right)\ge0\)\(\Leftrightarrow\left(2x-3\right)\left(1-x\right)\ge0\)

TH1: \(\hept{\begin{cases}2x-3\le0\\1-x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le\frac{3}{2}\\1\le x\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le\frac{3}{2}\\x\ge1\end{cases}}\Leftrightarrow1\le x\le\frac{3}{2}\)

TH2: \(\hept{\begin{cases}2x-3\ge0\\1-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge\frac{3}{2}\\1\ge x\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge\frac{3}{2}\\x\le1\end{cases}}\)( vô lý )

Vậy \(minA=1\Leftrightarrow1\le x\le\frac{3}{2}\)

=(x^2+y^2+2xy​)+(2x+2y)+3

=((x+y)² +2(x+y) +1)+2

=(x+y+1)²+2

vậy A_min=2

\(A=x^2+y^2+2xy+2x+2y+3\)

<=>\(A=x^2+2x\left(y+1\right)+y^2+2y+3\)

<=>\(A=x^2+2x\left(y+1\right)+\left(y^2+2y+1\right)+2\)

<=>\(A=x^2+2x\left(y+1\right)+\left(y+1\right)^2+2\)

<=>\(A=\left(x+y+1\right)^2+2\ge2\)

P=(x-1)(2x+3)

=2x²-2x+3x-3

=2x²+x-3

\(=2\left(x^2+\frac{1}{2}x+\frac{1}{16}-\frac{49}{16}\right)=2\left(x+\frac{1}{4}\right)^2-\frac{49}{8}\ge-\frac{49}{8}\)

dấu = xảy r khi x=-1/4

a) \(x^2+2x+3\)

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

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

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

Ta có:

\(\left(x+1\right)^2\ge0\) với mọi x

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

Vậy MinA = 2 khi

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

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

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

MIN A = 2 <=> X= -1 
MIN B = 7/4 <=> X = -1/2
MAX E = 10<=> X= 3 
MAX P = `<=> X= 1

Ta có :

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

                                        \(=2\left(x^2+\frac{1}{2}x-\frac{3}{2}\right)\) \(=2\left(x^2+\frac{1}{2}x+\frac{1}{16}-\frac{1}{16}-\frac{3}{2}\right)\)

                                          \(=2\left(x^2+\frac{1}{2}x+\frac{1}{16}-\frac{23}{16}\right)\)

                                           \(=2\left(x+\frac{1}{4}\right)^2-\frac{23}{8}\ge-\frac{23}{8},\)với mọi x

Vậy \(MIN_P=\frac{-23}{8}\) khi \(x+\frac{1}{4}=0\Leftrightarrow x=\frac{-1}{4}\)