Bài 1 : 

\(\left(x-2\right)^2-\left(x-3^2\right)=\left(x-2\right)^2-\left(x-9\right)\)

\(=x^2-4x+4-x+9=x^2-5x+13\)

Bài 2 : 

a, \(P=\frac{1-4x^2}{4x^2-4x+1}=\frac{\left(1-2x\right)\left(2x+1\right)}{\left(2x-1\right)^2}\)

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

b, Thay x = -4 ta được : 

\(\frac{-2.\left(-4\right)-1}{2.\left(-4\right)-1}=\frac{8-1}{-8-1}=-\frac{7}{9}\)

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

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

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

\(\Leftrightarrow x\left(2x-1\right)\left(3x+2\right)=0\Leftrightarrow x=0;\frac{1}{2};-\frac{2}{3}\)

ta có 

\(x^2\left(x-1\right)-4x^2+8x-4=0\)

\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-4\left(x-1\right)\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

Ta có: \(x^2\left(x-1\right)-4x^2+8x-4=0\)

\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\\left(x-2\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

a) ( 3 + x )² = x² + 6x + 9

b) ( 5 - x )³ = 125 - 75x + 45x² - x³

c) ( 2x - 1 )( x² - x + 3 ) = 2x³ - 2x² + 6x - x² + x - 3 = 2x³ - 3x² + 7x - 3

d) \(\frac{9}{x^2+3x}-\frac{3-x}{x}=\frac{9}{x\left(x+3\right)}+\frac{x-3}{x}\)

\(=\frac{9}{x\left(x+3\right)}+\frac{\left(x-3\right)\left(x+3\right)}{x\left(x+3\right)}\)

\(=\frac{9+x^2-9}{x\left(x+3\right)}=\frac{x^2}{x\left(x+3\right)}=\frac{x}{x+3}\)

a, \(\left(3+x\right)^2=9+6x+x^2\)

b, \(\left(5-x\right)^3=125-75x+15x^2-x^3\)

c, \(\left(2x-1\right)\left(x^2-x+3\right)=2x^3-2x^2+6x-x^2+x-3=2x^3-3x^2+7x-3\)

d, \(\frac{9}{x^2+3x}-\frac{3-x}{x}=\frac{9}{x\left(x+3\right)}-\frac{3-x}{x}=\frac{9}{x\left(x+3\right)}+\frac{\left(x-3\right)\left(x+3\right)}{x\left(x+3\right)}\)

\(=\frac{9+x^2-9}{x\left(x+3\right)}=\frac{x^2}{x\left(x+3\right)}\)

A B C D M N

ta có diện tích ADM \(=\frac{AD.AM}{2}=\frac{AD.AB}{4}=\frac{a.b}{4}\)

diện tích DMN \(=\frac{AD.NM}{2}=\frac{AD.MB}{4}=\frac{AD.AB}{8}=\frac{a.b}{8}\)

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

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

\(\Leftrightarrow\left(x-3\right)\left[x\left(x-3\right)-1\ne0\right]=0\)

\(\Leftrightarrow x=3\)

ta có 

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

\(\Leftrightarrow\left(x-3\right)\left[\left(x-3\right)^2-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\\left(x-3\right)^2-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=3\\\left(x-2\right)\left(x-4\right)=0\end{cases}}\)pt dưới \(\Leftrightarrow\orbr{\begin{cases}x=2\\x=4\end{cases}}\)

vậy x=2 hoặc x=3 hoặc x=4