Thử: 

Gọi 2/5 stn=1/6 sth = x

=> stn=5x/2 ; sth=6x

stn+sth=17x/2=51

x=51:17x2=6

Vậy stn=5x/2=15 ; sth=6x=36

Gọi st1 là x

Số t2 la : 51-x 

Theo đề bài ta có pt:   \(\frac{2}{5}x=\frac{1}{6}\left(51-x\right)\)

\(\Leftrightarrow\)                         \(\frac{2}{5}x=\frac{17}{2}-\frac{1}{6}x\)

\(\Leftrightarrow\)          \(\frac{2}{5}x+\frac{1}{6}x=\frac{17}{2}\)

\(\Leftrightarrow\)                      \(\frac{17}{30}x=\frac{17}{2}\)

\(\Leftrightarrow\)                              \(x=15\)

Vậy :...................................

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

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

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

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

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

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

Gọi số học sinh lớp 8A là a , ta có : 

* Hai lớp 8A và 8B có tổng cộng 94 học sinh 

=> lớp 8B có : 94 - x (học sinh) 

* 25% số học sinh lớp 8A đạt loại giỏi bằng : \(\frac{25x}{100}=\frac{x}{4}\)(học sinh) 

* 20% số học sinh lớp 8B đạt loại giỏi bằng : \(\frac{20\left(94-x\right)}{100}=\frac{94-x}{5}\)(học sinh) 

Suy ra : \(\frac{x}{4}+\frac{94-x}{5}=21\)

\(\Leftrightarrow5x+4\left(94-x\right)=420\)

\(\Leftrightarrow5x+376-4x=420\)

\(\Leftrightarrow x=420-376=44\)

Vậy lớp 8A có 44 học sinh 

       lớp 8B có 94 - 44 = 50 học sinh 

Đáp số:50 học sinh

DKXD :\(x\ne1,x\ne-2\)

\(\Rightarrow2\left(x+2\right)=\left(x+2\right)\left(x-1\right)+2x\left(x-1\right)\)

\(\Leftrightarrow2x+4=x^2+x-2+2x^2-2x\)

\(\Leftrightarrow2x+4-x^2-x+2-2x^2+2x=0\)

\(\Leftrightarrow3x-3x^2+6=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\-3x-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\left(\in dk\right)\\x=-1\left(\in dk\right)\end{cases}}\)

\(S=\left\{2,-1\right\}\)

\(x^3-4x=0\)

\(\Rightarrow x\left(x^2-4\right)=0\)

\(\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)

\(\Rightarrow\)x = 0 , x - 2 = 0 hoặc x + 2 = 0

\(\Rightarrow\)\(x=0,x=2,x=-2\)

\(x^3-4x=0\)

\(\Leftrightarrow x\left(x^2-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)

Vậy \(x\in\left\{-2;0;2\right\}\)

          \(\frac{x-3}{2}+2< \frac{2x+1}{3}\)

\(\Leftrightarrow\)\(\frac{3\left(x-3\right)}{6}+\frac{12}{6}< \frac{2\left(2x+1\right)}{6}\)

\(\Leftrightarrow\)\(3x-9+12< 4x+2\)

\(\Leftrightarrow\)\(3x-4x< 2+9-12\)

\(\Leftrightarrow\)\(-x< -1\)

\(\Leftrightarrow\)\(x>1\)

Vậy :....................................

\(\frac{x-3}{2}+2< \frac{2x+1}{3}\)

\(\Rightarrow\frac{3\left(x-3\right)}{6}+\frac{12}{6}< \frac{2\left(2x+1\right)}{6}\)

\(\Leftrightarrow3x-9+12< 4x+2\)

\(\Leftrightarrow3x-4x< 2+9-12\)

\(\Leftrightarrow-x< -1\)

\(\Leftrightarrow x>1\)

Vây.....

\(2\frac{2}{x-1}=1+\frac{2x}{x+2}\)              \(\left(x\ne1;x\ne-2\right)\)

\(\Rightarrow\frac{2\left(x-1\right)+2}{x-1}=\frac{\left(x+2\right)+2x}{x+2}\)\(\Rightarrow2x^2+4x=3x^2+2x-3x+2\)

\(\Rightarrow\frac{2x-2+2}{x-1}=\frac{x+2+2x}{x+2}\)

\(\Rightarrow\frac{2x}{x-1}=\frac{3x+2}{x+2}\)

\(\Rightarrow2x\left(x+2\right)=\left(x-1\right)\left(3x+2\right)\)

\(\Rightarrow2x^2+4x=x\left(3x+2\right)-1\left(3x+2\right)\)

\(\Rightarrow2x^2+4x=x\left(3x+2\right)-1\left(3x+2\right)\)

\(2\frac{2}{x-1}=1+\frac{2x}{x+2}\) ĐKXĐ: \(\hept{\begin{cases}x\ne1\\x\ne-2\end{cases}}\)

=> \(\frac{2\left(x-1\right)+2}{x-1}=\frac{x+2+2x}{x+2}\)

=> \(\frac{2\left(x-1+1\right)}{x-1}=\frac{x+2\left(x+1\right)}{x+2}\)

=> \(\frac{2x}{x-1}=\frac{x+2\left(x+1\right)}{x+2}\)

=> \(2x\left(x+2\right)=x+2\left(x+1\right)\left(x-1\right)\)

=> \(2x^2+4x=x+2\left(x^2-1\right)\)

=> \(2x^2+4x=x+2x^2-2\)

=> \(2x^2+4x-x-2x^2+2=0\)

=> \(3x+2=0\)

=> \(3x=-2\)

=> \(x=-\frac{2}{3}\)