a)-19

b)22

lập bảng xét dấu đi

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

A =|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  \(x=\dfrac{7}{5}\)

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

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=sqrt{x-2}+sqrt{6-x}(2<=x<=6)`
Áp dụng BĐT `sqrtA+sqrtB>=sqrt{A+B}`
`=>A>=sqrt{x-2+6-x}=2`
Dấu "=" `<=>x=2` hoặc `x=6`
Áp dụng BĐT bunhia
`=>A<=sqrt{2(x-2+6-x)}=2sqrt2`
Dấu "=" `<=>x=4`
`C=sqrt{1+x}+sqrt{8-x}(-1<=x<=8)`
Áp dụng BĐT `sqrtA+sqrtB>=sqrt{A+B}`
`=>A>=sqrt{1+x+8-x}=3`
Dấu "=" `<=>x=-1` hoặc `x=8`
Áp dụng BĐT bunhia
`=>A<=sqrt{2(1+x+8-x)}=3sqrt2`
Dấu "=" `<=>x=7/2`

`D=2sqrt{x+5}+sqrt{1-2x}(-5<=x<=1/2)`
`=sqrt{4x+20}+sqrt{1-2x}`
Áp dụng BĐT `sqrtA+sqrtB>=sqrt{A+B}`
`=>D>=sqrt{4x+20+1-2x}=sqrt{2x+21}`
Mà `x>=-5`
`=>D>=sqrt{-10+21}=sqrt{11}`
Dấu "=" `<=>x=-5`