C=(2-x)(x+4)

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

=\(-\left(x^2-2x+1\right)+9\)\(\le9\)

MaxC=9 Dấu "=" xảy ra khi x=1

1) \(P=-2x^2-12x=-2\left(x^2+6x+9\right)+18=-2\left(x+3\right)^2+18\le18\)

\(maxP=18\Leftrightarrow x=-3\)

2) \(Q=-5x^2+10x=-5\left(x^2-2x+1\right)+5=-5\left(x-1\right)^2+5\le5\)

\(maxQ=5\Leftrightarrow x=1\)

3) \(A=-3x^2+12x-6=-3\left(x^2-4x+4\right)+6=-3\left(x-2\right)^2+6\le6\)

\(maxA=6\Leftrightarrow x=2\)

4) \(B=-2x^2-24x+12=-2\left(x^2+12x+36\right)+84=-2\left(x+6\right)^2+84\le84\)

\(maxB=84\Leftrightarrow x=-6\)

\(A=x^2+3x-5=x^2+3x+\frac{9}{4}-\frac{29}{4}\)

\(=\left(x+\frac{3}{2}\right)^2-\frac{29}{4}\ge-\frac{29}{4}\)

Vậy \(A_{min}=-\frac{29}{4}\Leftrightarrow x+\frac{3}{2}=0\Leftrightarrow x=-\frac{3}{2}\)

a: ĐKXĐ: x∉{2;-2;0;3}

b: \(A = \left(\dfrac{4x}{x+2}+\dfrac{8x^2}{4-x^2}\right):\left(\dfrac{x-1}{x^2-2x}-\dfrac{2}{x}\right)\)

\(=\frac{4x\left(x-2\right)-8x^2}{\left(x+2\right)\left(x-2\right)}:\frac{x-1-2\left(x-2\right)}{x\left(x-2\right)}\)

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

\(=\frac{-4x^2-8x}{\left(x+2\right)}\cdot\frac{x}{-x+3}=\frac{-4x\left(x+2\right)}{x+2}\cdot\frac{x}{-x+3}=\frac{4x^2}{x-3}\)

c: \(x^2+2x=15\)

=>\(x^2+2x-15=0\)

=>(x+5)(x-3)=0

=>x=-5(nhận) hoặc x=3(loại)

Khi x=-5 thì \(A=\frac{4\cdot\left(-5\right)^2}{-5-3}=\frac{4\cdot25}{-8}=\frac{-100}{8}=-\frac{25}{2}\)

d: |A|>A

=>A<0

=>\(\frac{4x^2}{x-3}<0\)

=>x-3<0

=>x<3

Kết hợp ĐKXĐ, ta được: x<3 và x∉{-2;0;2}