A = \(11-10x-x^2\)

\(A=-\left(x^2+10x-11\right)\)

\(A=-\left(x^2+2x5+25-11-25\right)\)

\(A=-\left(x+5\right)^2+36\)

Dấu "=" xảy ra khi \(x+5=0\)\(\Leftrightarrow x=-5\)

Vậy Max A= 36 khi x = -5

B bn làm tương tự nha, k cho mình nha bn <3 

III:
A: var a:array[1..50]of real;

B: a[5]:=8;

C: var a:array[1..50]of integer;

D: readln(dem[2]);

\(C=\left(23-x\right)\left(3x+5\right)+13\)

\(=69x+115-3x^2-5x+13\)

\(=-3x^2+64x+128\)

\(=-3\left(x^2-\dfrac{64}{3}x+\dfrac{1024}{9}\right)+\dfrac{1408}{3}\)

\(=-3\left(x-\dfrac{32}{3}\right)^2+\dfrac{1408}{3}\le\dfrac{1408}{3}\)

Vậy \(Max_C=\dfrac{1408}{3}\)

Để \(C=\dfrac{1408}{3}\) thì \(x-\dfrac{32}{3}=0\Rightarrow x=\dfrac{32}{3}\)

d, \(D=\left(2-3x\right)\left(3x+5\right)-7\)

\(=6x+10-9x^2-15x-7\)

\(=-9x^2-9x+3\)

\(=-9\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{21}{4}\)

\(=-9\left(x-\dfrac{1}{2}\right)^2+\dfrac{21}{4}\le\dfrac{21}{4}\)

Vậy \(Max_D=\dfrac{21}{4}\) khi \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)

a) \(A=\dfrac{7}{3}\left(x^2+1\right)\)

Ta có:

\(x^2\ge0\forall x\\ \Rightarrow x^2+1\ge1\forall x\)

Để \(A=\dfrac{7}{3}\left(x^2+1\right)\) đạt GTNN thì \(x^2+1\) đạt GTNN

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

Thay \(x^2+1=1\) vào \(A=\dfrac{7}{3}\left(x^2+1\right)\) ta có:

\(A=\dfrac{7}{3}.1\\ A=\dfrac{7}{3}\)

Vậy \(Max_A=\dfrac{7}{3}\) tại \(x=0\)

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