Question

Giải các phương trình sau:

a) (x²+x)²+ 4(x²+x)=12

b) \(\dfrac{x+1}{2008}\)+\(\dfrac{x+2}{2007}\)+\(\dfrac{x+3}{2006}\)=\(\dfrac{x+4}{2005}\)+\(\dfrac{x+5}{2004}\)+\(\dfrac{x+6}{2003}\)

Answer 1

\(\dfrac{x+1}{2008}+\dfrac{x+2}{2007}+\dfrac{x+3}{2006}=\dfrac{x+4}{2005}+\dfrac{x+5}{2004}+\dfrac{x+6}{2003}\)

⇔\(\dfrac{x+1}{2008}+1+\dfrac{x+2}{2007}+1+\dfrac{x+3}{2006}+1=\dfrac{x+4}{2005}+1+\dfrac{x+5}{2004}+1+\dfrac{x+6}{2003}+1\)

⇔ \(\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}=\dfrac{x+2009}{2005}+\dfrac{x+2009}{2004}+\dfrac{x+2009}{2003}\)

⇔ \(\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}-\dfrac{x+2009}{2005}-\dfrac{x+2009}{2004}-\dfrac{x+2009}{2003}=0\)

⇔ \(\left(x+2009\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}-\dfrac{1}{2005}-\dfrac{1}{2004}-\dfrac{1}{2003}\right)=0\)

⇔ x+2009=0

⇔ x=-2009

vậy x=-2009 là nghiệm của pt

Answer 2

a) ( x² + x )² + 4( x² + x ) = 12

<=> ( x² + x )² + 4( x² + x ) + 4 - 16 = 0

<=> ( x² + x + 2)² - 16 = 0

<=> ( x² + x + 2 + 4)( x² + x + 2 - 4) = 0

<=> ( x² + x + 6 )( x² + x - 2) = 0

Do : x² + x + 6

= x² + 2.\(\dfrac{1}{2}x+\dfrac{1}{4}+6-\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\) ≥ \(\dfrac{23}{4}\) > 0 ∀x

=> x² + x - 2 = 0

<=> x² - x + 2x - 2 = 0

<=> x( x - 1) + 2( x - 1) = 0

<=> ( x - 1)( x + 2 ) = 0

<=> x = 1 hoặc : x = - 2

KL.....

b) Kuroba kaito làm rùi nhé