Đáp án: c) x = -1 d) x = -2004
Giải thích các bước giải:
Bài 2:
c)
$\frac{x+1}{10}$ + $\frac{x+1}{11}$ + $\frac{x+1}{12}$ = $\frac{x+1}{13}$ + $\frac{x+1}{14}$
(x + 1)($\frac{1}{10}$ + $\frac{1}{11}$ + $\frac{1}{12}$) = (x + 1)($\frac{1}{13}$ + $\frac{1}{14}$)
(x + 1)($\frac{1}{10}$ + $\frac{1}{11}$ + $\frac{1}{12}$ - $\frac{1}{13}$ - $\frac{1}{14}$) = 0
=> x + 1 = 0
=> x = -1
Vậy x = -1
d)
$\frac{x+4}{2000}$ + $\frac{x+3}{2001}$ = $\frac{x+2}{2002}$ + $\frac{x+1}{2003}$
$\frac{x+4}{2000}$ + 1 + $\frac{x+3}{2001}$ + 1 = $\frac{x+2}{2002}$ + 1 + $\frac{x+1}{2003}$ + 1
$\frac{x+4+2000}{2000}$ + $\frac{x+3+2001}{2001}$ = $\frac{x+2+2002}{2002}$ + $\frac{x+1+2003}{2003}$
$\frac{x+2004}{2000}$ + $\frac{x+2004}{2001}$ = $\frac{x+2004}{2002}$ + $\frac{x+2004}{2003}$
(x + 2004)($\frac{1}{2000}$ + $\frac{1}{2001}$) = (x + 2004)($\frac{1}{2002}$ + $\frac{1}{2003}$)
=> x + 2004 = 0
=> x = -2004
Vậy x = -2004
