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

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

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

Bài 1: 

Ta có |x-8| > 0 với mọi x

=>A=37-|x-8| > 37 với mọi x

Vậy GTLN của A=37 với x-8=0 =>x=8

Bài 2 tương tự nhé

Học tốt :))

a)-19

b)22

a: -x^2<=0

=>-x^2+1<=1

=>A<=1

Dấu = xảy ra khi x=0

b: (x+1)^2>=0

=>-2(x+1)^2<=0

=>B<=8

Dấu = xảy ra khi x=-1

\(A=0,5-\left|x-3,5\right|\le0,5\\ A_{max}=0,5\Leftrightarrow x-3,5=0\Leftrightarrow x=3,5\\ B=-\left|1,4-x\right|2=-2\left|1,4-x\right|\le0\\ B_{min}=0\Leftrightarrow1,4-x=0\Leftrightarrow x=1,4\)

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

\(=\left(x+1\right)^2+4\ge4\forall x\)

\(\Leftrightarrow\dfrac{5}{x^2+2x+5}\le\dfrac{5}{4}\forall x\)

Dấu '=' xảy ra khi x=-1

\(A=\left|x-2018\right|-\left|x-2017\right|\le\left|x-2018-x+2017\right|=\left|-1\right|=1\)

Dấu "=" xảy ra <=> (x-2018)(x-2017) > 0

<=> \(\left[{}\begin{matrix}x>2018\\x< 2017\end{matrix}\right.\)

Vậy MaxA = 1 <=> \(\left[{}\begin{matrix}x>2018\\x< 2017\end{matrix}\right.\)

A = | x − 2018 | − | x − 2017 | ≤ | x − 2018 − x + 2017 | = | − 1 | = 1 Dấu "=" xảy ra <=> (x-2018)(x-2017) > 0 <=> [ x > 2018 x < 2017 Vậy MaxA = 1 <=> [ x > 2018 x < 2017