x² - 5x = 0

<=> x( x - 5 ) = 0

<=> x = 0 hoặc x - 5 = 0

<=> x = 0 hoặc x = 5

x² - 7x + 10 = 0

<=> x² - 2x - 5x + 10 = 0

<=> x( x - 2 ) - 5( x - 2 ) = 0

<=> ( x - 2 )( x - 5 ) = 0

<=> x - 2 = 0 hoặc x - 5 = 0

<=> x = 2 hoặc x = 5

\(x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\Leftrightarrow x=0;5\)

\(x^2-7x+10=0\Leftrightarrow x^2-2x-5x+10=0\)

\(\Leftrightarrow x\left(x-2\right)-5\left(x-2\right)=0\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\Leftrightarrow x=5;2\)

5x( x - y ) + 12x - 12y 

= 5x( x - y ) + 12( x - y )

= ( x - y )( 5x + 12 )

x² + 2xy - 9 + y²

= ( x² + 2xy + y² ) - 9

= ( x + y )² - 3²

= ( x + y - 3 )( x + y + 3 )

\(5x\left(x-y\right)+12x-12y\)

\(=5x\left(x-y\right)+12\left(x-y\right)\)

\(=\left(x-y\right)\left(5x+12\right)\)

\(x^2+2xy-9+y^2\)

\(=\left(x^2+2xy+y^2\right)-9\)

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

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

\(B=\left(\frac{x}{x+1}+\frac{x-1}{x}\right)\div\left(\frac{x}{x+1}-\frac{x-1}{x}\right)\)

ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne-1\end{cases}}\)

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

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

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

Để B = 1 => 2x² - 1 = 1

=> 2x² - 1 - 1 = 0

=> 2x² - 2 = 0

=> 2( x² - 1 ) = 0

=> 2( x - 1 )( x + 1 ) = 0

=> x - 1 = 0 hoặc x + 1 = 0

=> x = 1 ( tm ) hoặc x = -1 ( ktm )

Vậy x = 1 thì B = 1

a, \(3x\left(2x-3\right)-6x\left(1+x\right)=5\)

\(\Leftrightarrow6x^2-9x-6x-6x^2=5\)

\(\Leftrightarrow-15x=5\Leftrightarrow x=-\frac{1}{3}\)

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

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

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

\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-6x^2+12x-12\right)-x^3-1=0\)

\(\Leftrightarrow-x^3+9x^2-9x+12=0\)

a,\(3x\left(2x-3\right)-6x\left(1+x\right)=5\)

\(\Leftrightarrow6x^2-9x-6x-6x^2=5\)

\(\Leftrightarrow-15x=5\)

\(\Leftrightarrow x=-\frac{1}{3}\)

Bài làm 

\(x^3-2x^2+3x-4x^2+8x-12\)

\(=x^3-6x^2+11x-12\)