Bạn tham khảo:
Đáp án:
a/ $S =$ { $23$ }
b/ $S =$ { $-2005$ }
c/ $S =$ { $300$ }
Giải thích các bước giải:
a/ $\dfrac{x-23}{24} + \dfrac{x-23}{25} = \dfrac{x-23}{26} + \dfrac{x-23}{27}$
$⇔ \dfrac{x-23}{24} + \dfrac{x-23}{25} - \dfrac{x-23}{26} - \dfrac{x-23}{27} = 0 $
$⇔ ( x - 23 )( \dfrac{1}{24} + \dfrac{1}{25} - \dfrac{1}{26} - \dfrac{1}{27} )= 0 $
Vì $\dfrac{1}{24} + \dfrac{1}{25} - \dfrac{1}{26} - \dfrac{1}{27} ≠ 0$
$⇔ x - 23 = 0 $
$⇔ x = 23$
$S =$ { $23$ }
b/ $\dfrac{x+1}{2004} + \dfrac{x+2}{2003} = \dfrac{x+3}{2002} + \dfrac{x+4}{2001}$
$⇔ \dfrac{x+1}{2004} + \dfrac{x+2}{2003}+2 = \dfrac{x+3}{2002} + \dfrac{x+4}{2001} +2$
$⇔ \dfrac{x+1}{2004} + \dfrac{x+2}{2003} + 1 + 1 = \dfrac{x+3}{2002} + \dfrac{x+4}{2001} + 1 + 1$
$⇔ (\dfrac{x+1}{2004}+1) + (\dfrac{x+2}{2003}+1) = (\dfrac{x+3}{2002}+1) +( \dfrac{x+4}{2001}+1)$
$⇔ \dfrac{x+2005}{2004} + \dfrac{x+2005}{2003} = \dfrac{x+2005}{2002} + \dfrac{x+2005}{2001}$
$⇔ \dfrac{x+2005}{2004} + \dfrac{x+2005}{2003} - \dfrac{x+2005}{2002} - \dfrac{x+2005}{2001} = 0$
$⇔ ( x + 2005 ) ( \dfrac{1}{2004} + \dfrac{1}{2003} - \dfrac{1}{2002} - \dfrac{1}{2001})= 0 $
Vì $ \dfrac{1}{2004} + \dfrac{1}{2003} - \dfrac{1}{2002} - \dfrac{1}{2001} ≠ 0$
$⇔ x + 2005 = 0 $
$⇔ x = -2005$
$S =$ { $-2005$ }
c/ $\dfrac{201-x}{99} + \dfrac{203-x}{97} + \dfrac{205-x}{95} + 3 = 0 $
$⇔ \dfrac{201-x}{99} + \dfrac{203-x}{97} + \dfrac{205-x}{95} + 1 + 1 + 1 = 0$
$⇔ (\dfrac{201-x}{99}+1) + (\dfrac{203-x}{97}+1) + (\dfrac{205-x}{95}+1) = 0 $
$⇔ \dfrac{201-x}{99} + \dfrac{203-x}{97} + \dfrac{205-x}{95} = 0 $
$⇔ \dfrac{300-x}{99} + \dfrac{300-x}{97} + \dfrac{300 - x}{95} = 0 $
$⇔ ( 300 -x )( \dfrac{1}{99} + \dfrac{1}{97} + \dfrac{1}{95} ) = 0 $
Vì $\dfrac{1}{99} + \dfrac{1}{97} + \dfrac{1}{95} ≠ 0$
$⇔ 300 - x = 0 $
$⇔ x = 300$
$S =$ { $300$ }
