\(\left(2x-1\right)^3-8\left(x-1\right)\left(x^2+x+1\right)+12x^2=2x+1\)

\(\Leftrightarrow8x^3-12x^2+6x-1-8\left(x^3-1\right)+12x^2-2x-1=0\)

\(\Leftrightarrow4x+6=0\)

\(\Leftrightarrow2\left(2x+3\right)=0\)

\(\Leftrightarrow2x=-3\)

\(\Leftrightarrow x=\frac{-3}{2}\)

đề sai nhé

phải là

8. (x-1) . (x²+x+1)

a)\(3x-\dfrac{2}{5}=0=>3x=\dfrac{2}{5}=>x=\dfrac{2}{15}\)

b)\(\left(x-3\right)\left(2x+8\right)=0=>\left[{}\begin{matrix}x-3=0\\2x=-8\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)

c)\(3x^2-x-4=0=>3x^2+3x-4x-4=0=>\left(3x-4\right)\left(x+1\right)=0\)

\(=>\left[{}\begin{matrix}3x=4\\x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-1\end{matrix}\right.\)

\(B=\left|x-7\right|+\left|x+8\right|\)

\(\Rightarrow B=\left|7-x\right|+\left|x+8\right|\)

\(\Rightarrow B\ge\left|7-x+x+8\right|\)

\(\Rightarrow B\ge\left|15\right|\)

\(\Rightarrow B\ge15\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-7\right)\left(x+8\right)\ge0\)

Vậy \(B_{min}=15\Leftrightarrow\left(x-7\right)\left(x+8\right)\ge0\)

Bài làm:

Ta có: \(\left\{4x-2\left(x-3\right)-3\left[x-3\left(4-2x\right)+8\right]\right\}.\left(-3x\right)\)

\(=\left[4x-2x+6-3\left(x-12+6x+8\right)\right].\left(-3x\right)\)

\(=\left(2x+6-3x+36-18x-24\right).\left(-3x\right)\)

\(=\left(-19x\right).\left(-3x\right)\)

\(=57x^2\)

Lời giải:

$\frac{x^3+8}{x^2-2x+1}.\frac{x^2+3x+2}{1-x^2}=\frac{(x^3+8)(x^2+3x+2)}{(x^2-2x+1)(1-x^2)}$

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

$=\frac{(x+2)^2(x^2-2x+4)}{-(x-1)^3}$
 

a  A=4x-x^2+3

      =(x-2)^2-1

     MIN A= -1 khi (x-2)^2=0

      x-2=0

      x=2

B=x-x^2

B=-x^2+x

-B=x^2-x

-B=(x-1/2)^2-1/4

B=-(x-1/2)^2+1/4

MAX B=1/4 khi -(x-1/2)^2=0

x-1/2=0

x=1/2

N=2x-2x^2-5

-N=2x^2-2x+5

-N=2(x^2-x+2)+1

-N=2{(x-1/2)^2+7/4}+1

-N=2(x-1/2)^2+7/2+1

-N=2(x-1/2)^2+9/2

N=-2(x-1/2)^2-9/2

MAX N=-9/2 khi -2(x-1/2)^2=0

x-1/2=0

x=1/2