Helo Dung :)

\(P=\left(\frac{x-4}{x^3-1}+\frac{1}{x-1}\right)\div\left(1-\frac{x-8}{x^2+x+1}\right)\)

ĐKXĐ : \(x\ne1\)

a) \(=\left(\frac{x-4}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right)\div\left(\frac{x^2+x+1}{x^2+x+1}-\frac{x-8}{x^2+x+1}\right)\)

\(=\left(\frac{x-4+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right)\div\left(\frac{x^2+x+1-x+8}{x^2+x+1}\right)\)

\(=\frac{x^2+2x-3}{\left(x-1\right)\left(x^2+x+1\right)}\div\frac{x^2+9}{x^2+x+1}\)

\(=\frac{x^2-x+3x-3}{\left(x-1\right)\left(x^2+x+1\right)}\times\frac{x^2+x+1}{x^2+9}\)

\(=\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x^2+9\right)}=\frac{x+3}{x^2+9}\)

b) \(x^2-3x+2=0\)

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

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

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

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

Với x = 2 =>\(P=\frac{2+3}{2^2+9}=\frac{5}{13}\)

c) Đến đây rồi sao ? Bậc của tử thấp hơn bậc của mẫu thì chia hết kiểu gì ? :D

@Qing: Biết chết liền, thế tao mới hỏi ==

( 3x - 1 )² - ( x + 5 )² = 0

⇔ [ ( 3x - 1 ) - ( x + 5 ) ][ ( 3x - 1 ) + ( x + 5 ) ] = 0

⇔ ( 3x - 1 - x - 5 )( 3x - 1 + x + 5 ) = 0

⇔ ( 2x - 6 )( 4x + 4 ) = 0

⇔ 2x - 6 = 0 hoặc 4x + 4 = 0

⇔ x = 3 hoặc x = -1

<=> ( 3x - 1 - x - 5 )( 3x - 1 + x + 5 ) = 0

<=> ( 2x - 6 )( 4x + 4) = 0 

=> 

+)  2x - 6 = 0 => x = 3

+) 4x +4 = 0 => x = -1

( x -2 ).( x + 2 ).( x +2 )²

= ( x - 2 ).(x + 2 )³

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

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

x(5-4+x)

x^2

x + 3 - 4x

= -3x + 3

Ta có : x² + 4x - y² + 4

= ( x² + 4x + 4 ) - y²

= ( x + 2 )² - y²

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

\(x^2+4x-y^2+4\)

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

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

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