giải phương trình bậc nhất x+1/2007+x+2/2006=x+2001/7+x+2002/6 04/07/2021 Bởi Aubrey giải phương trình bậc nhất x+1/2007+x+2/2006=x+2001/7+x+2002/6
$\dfrac{x+1}{2007} + \dfrac{x+2}{2006} = \dfrac{x+2001}{7} + \dfrac{x+2002}{6}$ $⇔ \dfrac{x+1}{2007} + \dfrac{x+2}{2006} + 2 = \dfrac{x+2001}{7} + \dfrac{x+2002}{6} + 2$ $⇔ \dfrac{x+1}{2007} + \dfrac{x+2}{2006} + 1 + 1 = \dfrac{x+2001}{7} + \dfrac{x+2002}{6} + 1 + 1$ $⇔ (\dfrac{x+1}{[2007}+1) + (\dfrac{x+2}{2006}+1) = (\dfrac{x+2001}{7}+1) + (\dfrac{x+2002}{6}+1)$ $⇔ \dfrac{x+2008}{2007} + \dfrac{x+2008}{2006} = \dfrac{x+2008}{7} + \dfrac{x+2008}{6}$ $⇔ \dfrac{x+2008}{2007} + \dfrac{x+2008}{2006} – \dfrac{x+2008}{7} – \dfrac{x+2008}{6} = 0$ $⇔ ( x + 2008 ) + ( \dfrac{1}{2007} + \dfrac{1}{2006} – \dfrac{1}{7} – \dfrac{1}{6} ) = 0 $ $⇔ x + 2008 = 0$ $⇔ x = -2008$ Bình luận
$\dfrac{x+1}{2007} + \dfrac{x+2}{2006} = \dfrac{x+2001}{7} + \dfrac{x+2002}{6}$
$⇔ \dfrac{x+1}{2007} + \dfrac{x+2}{2006} + 2 = \dfrac{x+2001}{7} + \dfrac{x+2002}{6} + 2$
$⇔ \dfrac{x+1}{2007} + \dfrac{x+2}{2006} + 1 + 1 = \dfrac{x+2001}{7} + \dfrac{x+2002}{6} + 1 + 1$
$⇔ (\dfrac{x+1}{[2007}+1) + (\dfrac{x+2}{2006}+1) = (\dfrac{x+2001}{7}+1) + (\dfrac{x+2002}{6}+1)$
$⇔ \dfrac{x+2008}{2007} + \dfrac{x+2008}{2006} = \dfrac{x+2008}{7} + \dfrac{x+2008}{6}$
$⇔ \dfrac{x+2008}{2007} + \dfrac{x+2008}{2006} – \dfrac{x+2008}{7} – \dfrac{x+2008}{6} = 0$
$⇔ ( x + 2008 ) + ( \dfrac{1}{2007} + \dfrac{1}{2006} – \dfrac{1}{7} – \dfrac{1}{6} ) = 0 $
$⇔ x + 2008 = 0$
$⇔ x = -2008$
Đáp án: Mình làm bài ở dưới ảnh nha
Giải thích các bước giải: