Question

tìm x:

a) 3x(2x-7)-(6x+1)(x-15)-2010=0

b) 2x(x-2012)-x+2012=0

c) (x+2y)(x^2-2xy+4y^2)-8y^3+27=0

d)x^3+x^2-2x-8=0

Answer 1

Ta có : 3x(2x - 7) - (6x + 1)(x - 15) - 2010 = 0

=> 6x² - 21x - (6x² + x - 90x - 15) - 2010 = 0

=> 6x² - 21x - 6x² + 89x + 15 - 2010 = 0

=> 68x - 1995 = 0

 ? 

b) 2x(x - 2012) - x + 2012 = 0

=> 2x(x - 2012) - (x - 2012) = 0

=> (x - 2012) (2x - 1) = 0

⇔[

x−2012=0

2x−1=0

⇔[

x=2012

2x=1

⇔[

x=2012

x=12

Vậy x = {2012;12 }

Answer 2

Ta có : 3x(2x - 7) - (6x + 1)(x - 15) - 2010 = 0

=> 6x² - 21x - (6x² + x - 90x - 15) - 2010 = 0

=> 6x² - 21x - 6x² + 89x + 15 - 2010 = 0

=> 68x - 1995 = 0

 ? 

b) 2x(x - 2012) - x + 2012 = 0

=> 2x(x - 2012) - (x - 2012) = 0

=> (x - 2012) (2x - 1) = 0

\(\Leftrightarrow\orbr{\begin{cases}x-2012=0\\2x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2012\\2x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2012\\x=\frac{1}{2}\end{cases}}\)

Vậy x = \(\left\{2012;\frac{1}{2}\right\}\)