\(M=-x^2+12x+8=-\left(x-6\right)^2+44\le44\)

\(M_{max}=44\) khi \(x=6\)

\(N=a^2+9b^2+5a-6b=\left(a+\dfrac{5}{2}\right)^2+\left(3b-1\right)^2-\dfrac{41}{4}\ge-\dfrac{41}{4}\)

\(N_{min}=-\dfrac{41}{4}\) khi \(\left(a;b\right)=\left(-\dfrac{5}{2};\dfrac{1}{3}\right)\)

\(Q=3\left(a-5\right)^2-82\ge-82\)

\(Q_{min}=-82\) khi \(a=5\)

\(A=-9x^2-12x+4\)

\(=-\left[\left(3x\right)^2+2\times3x\times2+2^2-2^2-4\right]\)

\(=-\left[\left(3x+2\right)^2-8\right]\)

\(\left(3x+2\right)^2\ge0\)

\(\left(3x+2\right)^2-8\ge-8\)

\(-\left[\left(3x+2\right)^2-8\right]\le8\)

Vậy Max A = 8 khi x = \(-\frac{2}{3}\)

\(A=-9x^2-12x+4=-\left(9x^2+12x-4\right)=-\left[\left(3x\right)^2+2.2.3x+2^2-8\right]\)

\(=-\left[\left(3x+2\right)^2-8\right]=-\left(3x+2\right)^2+8\)

Do \(\left(3x+2\right)^2\ge0\Rightarrow-\left(3x+2\right)^2\le0\Rightarrow-\left(3x+2\right)^2+8\le8\)

Đẳng thức xảy ra khi: \(3x+2=0\Rightarrow x=\frac{-2}{3}\)

Vậy giá trị lớn nhất của \(-9x^2-12x+4\)là 8 khi \(x=\frac{-2}{3}\)

a) \(4x^2+12x+1=\left(4x^2+12x+9\right)-8=\left(2x+3\right)^2-8\ge-8\)

\(ĐTXR\Leftrightarrow x=-\dfrac{3}{2}\)

b) \(4x^2-3x+10=\left(4x^2-3x+\dfrac{9}{16}\right)+\dfrac{151}{16}=\left(2x-\dfrac{3}{4}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\)

\(ĐTXR\Leftrightarrow x=\dfrac{3}{8}\)

c) \(2x^2+5x+10=\left(2x^2+5x+\dfrac{25}{8}\right)+\dfrac{55}{8}=\left(\sqrt{2}x+\dfrac{5\sqrt{2}}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\)

\(ĐTXR\Leftrightarrow x=-\dfrac{5}{4}\)

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

\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)

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

\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)

f) \(4x^2+2y^2+4xy+4y+5=\left(4x^2+4xy+y^2\right)+\left(y^2+4y+4\right)+1=\left(2x+y\right)^2+\left(y+2\right)^2+1\ge1\)

\(ĐTXR\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

a: Ta có: \(4x^2+12x+1\)

\(=4x^2+12x+9-8\)

\(=\left(2x+3\right)^2-8\ge-8\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)

b: Ta có: \(4x^2-3x+10\)

\(=4\left(x^2-\dfrac{3}{4}x+\dfrac{5}{2}\right)\)

\(=4\left(x^2-2\cdot x\cdot\dfrac{3}{8}+\dfrac{9}{64}+\dfrac{151}{64}\right)\)

\(=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{8}\)

c: Ta có: \(2x^2+5x+10\)

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

\(=2\left(x^2+2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{55}{16}\right)\)

\(=2\left(x+\dfrac{5}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{5}{4}\)

a: Ta có: \(A=x^2+3x+4\)

\(=x^2+2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{7}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)

a: Ta có: \(A=2x^2+12x+11\)

\(=2\left(x^2+6x+\dfrac{11}{2}\right)\)

\(=2\left(x^2+6x+9-\dfrac{7}{2}\right)\)

\(=2\left(x+3\right)^2-7\ge-7\forall x\)

Dấu '=' xảy ra khi x=-3

\(A=2\left(x^2+6x+36\right)-61=2\left(x+6\right)^2-61\ge-61\\ A_{min}=-61\Leftrightarrow x=-6\\ B=-\left(x^2-18x+81\right)+100=-\left(x-9\right)^2+100\le100\\ B_{max}=100\Leftrightarrow x=9\)

A=3(x^2+2/3x-1)

=3(x^2+2*x*1/3+1/9-10/9)

=3(x+1/3)^2-10/3>=-10/3

Dấu = xảy ra khi x=-1/3

\(B=1+\dfrac{15}{x^2+x+5}=1+\dfrac{15}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}}< =1+15:\dfrac{19}{4}=1+\dfrac{60}{19}=\dfrac{79}{19}\)

Dấu = xảy ra khi x=-1/2