\(\frac{2013.2014-2011}{2014.2011+2017}\)=\(\frac{2014\left(2011+2\right)-2011}{2014.2011+2017}\)

                                        =\(\frac{2011.2014+2017}{2011.2014+2017}\)

                                       =1

28.

\(\Leftrightarrow2x+1=x-1\)

\(\Leftrightarrow x=-2\)

29.

\(\Leftrightarrow x^3-2x^2-3x^2+6x=0\)

\(\Leftrightarrow x^2\left(x-2\right)-3x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2-3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x^2-3x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x\left(x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x=3\end{matrix}\right.\)

30.\(\Leftrightarrow3x^2+3x+2x+2=0\)

\(\Leftrightarrow3x\left(x+1\right)+2\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{2}{3}\end{matrix}\right.\)

\(28,\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=x-1\\2x+1=-x+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\\ 29,\\ \Leftrightarrow x\left(x^2-5x+6\right)=0\\ \Leftrightarrow x\left(x^2-2x-3x+6\right)=0\\ \Leftrightarrow x\left[x\left(x-2\right)-3\left(x-2\right)\right]=0\\ \Leftrightarrow x\left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\) 

\(30,\\ \Leftrightarrow3x^2+3x+2x+2=0\\ \Leftrightarrow3x\left(x+1\right)+2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2}{3}\\x=-1\end{matrix}\right.\)

\(\Rightarrow12+20x-60=45x-15\Leftrightarrow25x=-33\Leftrightarrow x=-\dfrac{33}{25}\)

12+20x−60=45x−15

25x=−33

x=−\(\dfrac{33}{25}\)

a) \(A=9x^2+5x+1\)

\(A=9x^2+5x+\frac{25}{36}+\frac{11}{36}\)

\(A=\left(3x+\frac{5}{6}\right)^2+\frac{11}{36}\) 

Có:  \(\left(3x+\frac{5}{6}\right)^2\ge0\)

\(\Rightarrow\left(3x+\frac{5}{6}\right)^2+\frac{11}{36}\ge\frac{11}{36}\)

Dấu = xảy ra khi: \(\left(3x+\frac{5}{6}\right)^2=0\Rightarrow3x+\frac{5}{6}=0\)

\(\Rightarrow x=-\frac{5}{18}\)

Vậy: \(Min_A=\frac{11}{36}\) tại \(x=-\frac{5}{18}\)

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

\(B=4x^2+12x+9-17\)

\(B=\left(2x+3\right)^2-17\)

Có: \(\left(2x+3\right)^2\ge0\)

\(\Rightarrow\left(2x+3\right)^2-17\ge-17\)

Dấu = xảy ra khi: \(\left(2x+3\right)^2=0\Rightarrow2x+3=0\)

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

Vậy: \(Min_B=-17\) tại \(x=-\frac{3}{2}\)

mình chịu

^x2-3.(x-1)

(x-1)²

^=>x2-3

x-1

a. 2x².(3x³ + 2x)  

= 2x².3x³ + 2x².2x 

= 6x⁵ + 4x³

b. 3x.(x² + 2x + 2) 

= 3x.x² + 3x.2x + 3x.2 

= 3x³ + 6x² + 6x 

`a)2x^2.(3x^3+2x)`

`=2x^2 .3x^3+2x^2 .2x`

`=6x^5+4x^3`

`b)3x(x^2+2x+2)`

`=3x.x^2+3x.2x+3x.2`

`=3x^3+6x^2+6x`