a: Ta có: \(\left(8x^2-4x\right):\left(-4x\right)-\left(x+2\right)=8\)

\(\Leftrightarrow-2x+1-x-2=8\)

\(\Leftrightarrow-3x=9\)

hay x=-3

b: Ta có: \(\left(2x^4-3x^3+x^2\right):\left(-\dfrac{1}{2}x^2\right)+4\left(x-1\right)^2=0\)

\(\Leftrightarrow-4x^2+6x-2+4x^2-8x+4=0\)

\(\Leftrightarrow-2x=-2\)

hay x=1

a: \(-x^2+2x-4\)

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

\(=-\left(x^2-2x+1+3\right)\)

\(=-\left\lbrack\left(x-1\right)^2+3\right\rbrack=-\left(x-1\right)^2-3\le-3\forall x\)

=>\(\frac{1}{-x^2+2x-4}\ge-\frac13\forall x\)

Dấu '=' xảy ra khi x-1=0

=>x=1

b: \(-4x^2+12x-13\)

\(=-\left(4x^2-12x+13\right)\)

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

\(=-\left\lbrack\left(2x-3\right)^2+4\right\rbrack=-\left(2x-3\right)^2-4\le-4\forall x\)

=>\(\frac{12}{-4x^2+12x-13}\ge\frac{12}{-4}=-3\forall x\)

Dấu '=' xảy ra khi 2x-3=0

=>2x=3

=>\(x=\frac32\)

c: Đặt \(A=\frac{x^2-4x-4}{x^2-4x+5}\)

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

\(=1-\frac{9}{x^2-4x+5}\)

Ta có: \(x^2-4x+5\)

\(=x^2-4x+4+1\)

\(=\left(x-2\right)^2+1\ge1\forall x\)

=>\(\frac{9}{\left(x-2\right)^2+1}\le\frac91=9\forall x\)

=>\(-\frac{9}{\left(x-2\right)^2+1}\ge-9\forall x\)

=>\(A=-\frac{9}{\left(x-2\right)^2+1}+1\ge-9+1=-8\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2

e: Đặt \(B=\frac{x^2-2011}{4\left(x^2+1\right)}\)

\(=\frac14\cdot\frac{4x^2-8044}{4x^2+4}=\frac14\cdot\frac{x^2-2011}{x^2+1}=\frac14\left(\frac{x^2+1-2012}{x^2+1}\right)=\frac14\left(1-\frac{2012}{x^2+1}\right)\)

Ta có: \(x^2+1\ge1\forall x\)

=>\(\frac{2012}{x^2+1}\le2012\forall x\)

=>\(-\frac{2012}{x^2+1}\ge-2012\forall x\)

=>\(1-\frac{2012}{x^2+1}\ge-2012+1=-2011\forall x\)

=>\(\frac14\left(1-\frac{2012}{x^2+1}\right)\ge-\frac{2011}{4}\forall x\)

Dấu '=' xảy ra khi x=0

`2//(5x-8)-3(4x-5)=4(3x-4)`

`<=>5x-8-12x+15=12x-16`

`<=>-19x=-23`

`<=>x=23/19`     Vậy `x=23/19`

`3//2(x^3-1)-2x^2(x+2x^4)+(4x^5+4)x=6`

`<=>2x^3-2-2x^3-4x^6+4x^6+4x=6`

`<=>4x=8`

`<=>x=2`     Vậy `x=2`

`A`

\(1,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 2,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 3,\left(x+1\right)^2+2\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x+1+2\right)=0\\ \Rightarrow\left(x+1\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

x²+4x+4=0
(x+2)²=0
x+2=0
x=+-2
câu 1 giống câu 2
(x+1)²+2(x+1)=0
(x+1+2)(x+1)=0
Th1: x+3=0           Th2: x+1=0
            x=-3                      x=-1
vậy ...

2:

=>x^3-1-2x^3-4x^6+4x^6+4x=6

=>-x^3+4x-7=0

=>x=-2,59

4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50

=>-62x+12=-50

=>x=1