\(\text{ C = 3 - | x + 2 |}\)

               \(\left|x+2\right|\ge0\)

\(\Rightarrow3-\left|x+2\right|\ge3-0\)

\(\Rightarrow3-\left|x+2\right|\ge3\)

\(\Rightarrow C\ge3\)

\(\Rightarrow C=3\Leftrightarrow\left|x+2\right|=0\)

                    \(\Rightarrow x+2=0\)

                     \(\Rightarrow x=0-2\)

                     \(\Rightarrow x=-2\)

Vậy \(\text{Max C = 3 }\Leftrightarrow x=-2\)

\(!x+2!\ge0\Leftrightarrow3-!x+2!\le3\)

"=" xảy ra khi x=-2

\(!3x-15!\ge0\)

\(!3x-15!+8\ge8\)

dấu = xảy ra khi x=5

\(A=\left|3x-15\right|+8\)

            \(\left|3x-15\right|\ge0\)

\(\Rightarrow\left|3x-15\right|+8\ge0+8\)

\(\Rightarrow\left|3x-15\right|+8\ge8\)

\(\Rightarrow A\ge8\)

\(\Rightarrow A=8\Leftrightarrow\left|3x-15\right|=0\)

                   \(\Rightarrow3x-15=0\)

                     \(\Rightarrow x=\left(0+15\right):3=5\)

Vậy \(MinA=8\Leftrightarrow x=5\)

B = -3(x² +3x + 9/4 -9/4) -7 

B = -3(x+3/2)² -7 +27/4

GTLN B = -1/4

B = -3(x² +3x + 9/4 -9/4) -7 

B = -3(x+3/2)² -7 +27/4

GTLN B = -1/4

\(A=\frac{1}{\left|x-2\right|+3}\)

Để x đạt giá trị lớn nhất thì \(\left|x-2\right|+3\) đạt giá trị nhỏ nhất

Có: \(\left|x-2\right|\ge0\Rightarrow\left|x-2\right|+3\ge3\)

Dấu bằng xảy ra khi \(x-2=0\Rightarrow x=2\)

Vậy \(Max_A=\frac{1}{3}\)tại \(x=2\)

\(A=\frac{1}{\left|x+2\right|}+3\)Trường hợp : \(x+2\ne0\Rightarrow x=-2\)

Ta có : \(\left|x+2\right|>0\Rightarrow\frac{1}{\left|x+2\right|}>0\)

\(\Rightarrow A=\frac{1}{\left|x+2\right|}+3\ge3\)

MAx \(A=3\Leftrightarrow\frac{1}{\left|x+2\right|}=0\left(vôlys\right)\)

Vậy A ko tồn tại giá trị lớn nhất

a:Ta có: \(A=-4x^2+x-1\)

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

\(=-4\left(x^2-2\cdot x\cdot\dfrac{1}{8}+\dfrac{1}{64}+\dfrac{63}{64}\right)\)

\(=-4\left(x-\dfrac{1}{8}\right)^2-\dfrac{63}{16}\le-\dfrac{63}{16}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{8}\)

b: Ta có: \(B=-3x^2+5x+6\)

\(=-3\left(x^2-\dfrac{5}{3}x-2\right)\)

\(=-3\left(x^2-2\cdot x\cdot\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{97}{36}\right)\)

\(=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{12}\le\dfrac{97}{12}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{6}\)

c: Ta có: \(C=-x^2+3x+4\)

\(=-\left(x^2-3x-4\right)\)

\(=-\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{25}{4}\right)\)

\(=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)

\(M=4x-x^2+3\)

\(=-x^2+4x+3\)

\(=-x^2+4x-4+7\)

\(=-\left(x^2-4x+4\right)+7\)

\(=-\left(x-2\right)^2+7\)

Vì; \(-\left(x-2\right)^2+7\le7\forall x\)

=> Max M =7 tại \(-\left(x-2\right)^2=0\Rightarrow x=2\)

Ta có: \(N=x-x^2=-x^2+x\)

\(=-x^2+x-\frac{1}{4}+\frac{1}{4}\)

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

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

Vì: \(-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x\)

=> Max N =1/4 tại \(-\left(x-\frac{1}{2}\right)^2=0\Rightarrow x=\frac{1}{2}\)

=.= hok tốt!!

a)\(M=4x-x^2+3\)

\(M=-x^2+4x+3\)

\(M=-x^2+4x-4+7\)

\(M=-\left(x-2\right)^2+7\le7.Với\forall x\in Q\)

Dấu "=" xảy ra khi x = 2

Vậy Max M = 7 <=> x = 2

b)\(N=x-x^2=-x^2+x\le x\)

Dấu "=" xảy ra khi x = 0

=> Max N = 0 <=> x = 0