Gọi vận tốc người thứ hai đi từ B -> A là V, thời gian người thứ 2 đi từ B đến khi gặp người thứ nhất là T 

=> vận tốc người thứ nhất đi từ A -> B là 3/4.V, thời gian thứ 1 đi từ A đến khi gặp người thứ hai là 2/5.T

Theo bài ra ta có: \(\frac{3}{4}V\cdot\frac{2}{5}T+V.T=16,5\)

\(\Leftrightarrow\frac{3}{10}.V.T+V.T=16,5\)

\(\Leftrightarrow\frac{13}{10}.V.T=\frac{165}{10}\)

\(\Leftrightarrow V.T=\frac{165}{10}\cdot\frac{10}{13}=\frac{165}{13}\left(\frac{km}{h}\right)\)

\(\Leftrightarrow\frac{3}{10}V.T=\frac{165}{13}\cdot\frac{3}{10}=\frac{495}{10}\left(\frac{km}{h}\right)\)

Vậy ... 

(x - 2)(x + 2) - x(x - 3) = 9

<=> x² - 4 - x² + 3x = 9

<=> -4 + 3x = 9

<=> 3x - 4 = 9

<=> 3x = 9 + 4

<=> 3x = 13

<=> x = 13 : 3

=> x = 13/3

\(\text{A)}\left(x+2y\right)^2-\left(x-2y\right)^2=\left(x+2y+x-2y\right)\left(x+2y-x+2y\right)=2x.4y=8xy\)

\(\text{B)}\left(2x+3\right)^2-2\left(4x^2-9\right)+\left(2x-3\right)^2=\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-3\right)+\left(2x-3\right)^2\)

\(=\left(2x+3-2x+3\right)^2=6^2=36\)

a) ( x + 2y )² - ( x - 2y )²

= ( x + 2y + x - 2y ).( x + 2y - x + 2y )

= 2x.4y

= 8xy

b) ( 2x + 3 )² - 2.( 4x² - 9 ) + ( 2x - 3 )²

= ( 2x + 3 )² - 2.[ ( 2x )² - 3² ] + ( 2x - 3 )²

= ( 2x + 3 )² - 2.( 2x + 3 ).( 2x - 3 ) + ( 2x - 3 )²

= ( 2x + 3 - 2x + 3 )²

= 6²

= 36

\(\left(5x+1\right)^2=\left(2x-3\right)^2\)

\(\Rightarrow\left(5x+1\right)^2-\left(2x-3\right)^2=0\)

\(\Rightarrow\left(5x+1+2x-3\right)\left(5x+1-2x+3\right)=0\)

\(\Rightarrow\left(7x-2\right)\left(3x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}7x-2=0\\3x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{2}{7}\\x=\frac{-4}{3}\end{cases}}}\)

Vậy.......

\(a,x^3-4x^2+8x-8\)

\(=\left(x^3-8\right)-\left(4x^2-8x\right)\)

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

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

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

\(=\left(x-2\right)\left(x^2+3x+6\right)\)

\(b,\left(2xy+1\right)^2-\left(2x+y\right)^2\)

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

\(=\left[\left(2xy+2x\right)+\left(y+1\right)\right]\cdot\left[\left(2xy-2x\right)-\left(y-1\right)\right]\)

\(=\left[2x\left(y+1\right)+1\cdot\left(y+1\right)\right]\cdot\left[2x\left(y-1\right)-1\cdot\left(y-1\right)\right]\)

\(=\left[\left(y+1\right)\left(2x+1\right)\right]\cdot\left[\left(y-1\right)\left(2x-1\right)\right]\)     

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

\(=\left(y^2-1\right)\left(4x^2-1\right)\)

\(c,1+6x-6x^2-x^3\)

\(=\left(1-x^3\right)-\left(6x^2-6x\right)\)

\(=\left(1^3-x^3\right)-6x\left(x-1\right)\)

\(=\left(1-x\right)\left(1+2x+x^2\right)+6x\left(1-x\right)\)

\(=\left(1-x\right)\left(1+2x+x^2+6x\right)\)

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

\(a,x^2-9-2\left(x+3\right)^2=0\)

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

\(\Rightarrow\left(x-3\right)\left(x+3\right)-2\left(x+3\right)\left(x+3\right)=0\)

\(\Rightarrow\left(x+3\right)\left[x-3-2\cdot\left(x+3\right)\right]=0\)

\(\Rightarrow\left(x+3\right)\left[x-3-2x-6\right]=0\)

\(\Rightarrow\left(x+3\right)\left(-x-9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\-x-9=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\-x=9\Rightarrow x=-9\end{cases}}\)

\(b,\left(5x+1\right)^2=\left(2x-3\right)^2\)

\(\Rightarrow\left(5x+1\right)^2\div\left(2x-3\right)^2=0\)

\(\Rightarrow\left[\left(5x+1\right)\div\left(2x-3\right)\right]^2=0\)

\(\Rightarrow\left(5x+1\right)\div\left(2x-3\right)=0\)

\(\Rightarrow5x+1=0\)

\(\Rightarrow x=-\frac{1}{5}\)

\(2xy-x^2+3y^2-4y+1\)

\(=-\left(x^2-2xy+y^2\right)+4y^2-4y+1\)

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

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

\(=\left(y+x-1\right)\left(3y-x-1\right)\)