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

\(a.A=\left(x-2\right)^2+\left(y+1\right)^2+1\ge1\forall x;y\) . " = " \(\Leftrightarrow x=2;y=-1\) 

b.\(B=7-\left(x+3\right)^2\le7\forall x\)  " = " \(\Leftrightarrow x=-3\)

c.\(C=\left|2x-3\right|-13\ge-13\forall x\)  " = " \(\Leftrightarrow x=\dfrac{3}{2}\)

d.\(D=11-\left|2x-13\right|\le11\forall x\)  " = " \(\Leftrightarrow x=\dfrac{13}{2}\)

A =|3x-4| + |5x-7| -x +2025

- Nếu x < \(\dfrac{4}{3}\):

\(\Rightarrow\) \(\left\{{}\begin{matrix}3x-4< 0\\5x-7< 0\end{matrix}\right.\) \(\Rightarrow\) \(\left\{{}\begin{matrix}\text{|}3x-4\text{|}=-3+4\\\text{|}5x-7\text{|}=-5x+7\end{matrix}\right.\) 

\(\Rightarrow\) \(A=-3x+4-5x+7-x+2025\) 

Vì x \(< \dfrac{4}{3}\) \(\Rightarrow\) \(9x< 12\) \(\Rightarrow\) \(-9x>-12\) 

\(\Rightarrow\) \(-9x+2036>2024\) 

\(\Rightarrow\) A \(>2024\) ( Loại)

Nếu \(\dfrac{4}{3}\) \(\le\) x \(< \dfrac{7}{5}\) 

\(\Rightarrow\) \(\left\{{}\begin{matrix}3x-4>0\\5x-7< 0\end{matrix}\right.\) \(\Rightarrow\) \(\left\{{}\begin{matrix}\text{|}3x-4\text{|}=3x-4\\\text{|}5x-7\text{|}=-5x+7\end{matrix}\right.\) 

\(\Rightarrow\) A= \(-3x-4-5x+7-x+2025\) 

       =   \(-3x+2028\) 

Ta có: \(\dfrac{4}{3}\) \(\le x\) \(\Rightarrow\) \(-3x\) \(>\dfrac{-21}{5}\) 

\(\Rightarrow\) 2024 \(\ge\) \(-3x+2028>\dfrac{10119}{5}\) ( loại)

Nếu x :

\(\ge\dfrac{7}{5}\\ \Rightarrow\left\{{}\begin{matrix}3x-4>0\\5x-7>0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\text{|}3x-4\text{|}=3x-4\\\text{|}5x-7\text{|}=5x-7\end{matrix}\right.\\ \Rightarrow A=3x-4+5x-7-x+2025\) 

  \(=7x+2014\) 

Vì \(x\ge\dfrac{7}{5}\) \(\Rightarrow\) \(7x\ge\dfrac{49}{5}\) 

\(\Rightarrow\) \(7x+2014\) \(\ge\dfrac{19}{5}+2014=\dfrac{10119}{5}\) 

\(\Rightarrow\) A \(\ge\) \(\dfrac{10119}{5}\) (  t/m)

Vậy A đạt GTNN khi A bằng \(\dfrac{10119}{5}\)

Dấu "=" xảy ra khi  \(x=\dfrac{7}{5}\)

a)-19

b)22

giải nhanh đi nhé mik cần gấp ai lm đủ đúng hết mik k mun cho nha giải đủ các bước nhé cảm ưn các bạn trước giúp mik nha^.^><hihiii

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

vi \(\left(x+1\right)^2\ge0\)(voi moi x)

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

Vay GTNN cua A =2 khi x=-1

2)  Goi 2 so nguyen lien tiep do la x va x+1

TDTC x+1-x=1

Vi 1 la so le nen x+1-x la so le 

Vay .......

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

\(=-2y\cdot2x=-4xy\)(dpcm)

4) \(Q=-x^2+6x+1=-\left(x^2-6x-1\right)=-\left(x^2-6x+9-10\right)=-\left(x-3\right)^2+10\)

Vi \(\left(x-3\right)^2\ge0\)(voi moi x)

\(\Rightarrow-\left(x-3\right)^2\le0\)(voi moi x)

\(\Rightarrow-\left(x-3\right)^2+10\le10\)(voi moi x)

Vay GTLN cua Q=10 khi x=3

\(a,=x^2-8x+16+1=\left(x-4\right)^2+1\ge1\)

Dấu \("="\Leftrightarrow x=4\)

\(b,=\left(4x^2-12x+9\right)+4=\left(2x-3\right)^2+4\ge4\)

Dấu \("="\Leftrightarrow x=\dfrac{3}{2}\)

\(c,=\left(9x^2-2\cdot3\cdot\dfrac{1}{3}x+\dfrac{1}{9}\right)+\dfrac{26}{9}=\left(3x-\dfrac{1}{3}\right)^2+\dfrac{26}{9}\ge\dfrac{26}{9}\)

Dấu \("="\Leftrightarrow3x=\dfrac{1}{3}\Leftrightarrow x=\dfrac{1}{9}\)