Bạn tham khảo:
Đáp án:
a/ $S =$ { $-2005$ }
b/ $S =$ { $2000$ }
Giải thích các bước giải:
a/ $\dfrac{x+6}{1999} + \dfrac{x+8}{1997} = \dfrac{x+10}{1995} + \dfrac{x+12}{1993}$
$⇔ \dfrac{x+6}{1999} + \dfrac{x+8}{1997} + 2 = \dfrac{x+10}{1995} + \dfrac{x+12}{1993} + 2$
$⇔ \dfrac{x+6}{1999} + \dfrac{x+8}{1997} + 1 + 1= \dfrac{x+10}{1995} + \dfrac{x+12}{1993} + 1 + 1$
$⇔ (\dfrac{x+6}{1999}+1) + (\dfrac{x+8}{1997}+1) = (\dfrac{x+10}{1995}+1) + (\dfrac{x+12}{1993}+1)$
$⇔ \dfrac{x+2005}{1999} + \dfrac{x+2005}{1997} - \dfrac{x+2005}{1995} - \dfrac{x+2005}{1993} = 0 $
$⇔ ( x + 2005 ) (\dfrac{1}{1999} + \dfrac{1}{1997} - \dfrac{1}{1995} - \dfrac{1}{1993} ) = 0 $
Vì $\dfrac{1}{1999} + \dfrac{1}{1997} - \dfrac{1}{1995} - \dfrac{1}{1993} ≠ 0$
$⇔ x + 2005 = 0 $
$⇔ x = -2005$
$S =$ { $-2005$ }
b/ $\dfrac{1909-x}{91} + \dfrac{1907-x}{93} + \dfrac{1905-x}{95} + \dfrac{1903-x}{91} + 4 = 0 $
$⇔ \dfrac{1909-x}{91} + \dfrac{1907-x}{93} + \dfrac{1905-x}{95} + \dfrac{1903-x}{91} + 1 + 1 + 1 + 1 = 0$
$⇔ (\dfrac{1909-x}{91}+1) + (\dfrac{1907-x}{93}+1) + (\dfrac{1905-x}{95}+1) + (\dfrac{1903-x}{91} +1) = 0$
$⇔ \dfrac{2000-x}{91} + \dfrac{2000-x}{93} + \dfrac{2000-x}{95} + \dfrac{2000-x}{91} = 0 $
$⇔ ( 2000 - x )( \dfrac{1}{91} + \dfrac{1}{93} + \dfrac{1}{95} + \dfrac{1}{91} ) = 0 $
Vì $\dfrac{1}{91} + \dfrac{1}{93} + \dfrac{1}{95} + \dfrac{1}{91} ≠ 0$
$⇔ 2000 -x =0 $
$⇔ x = 2000$
$S =$ {$2000$ }
