$\frac{1}{(x+1)(x+2)}$ + $\frac{1}{(x+2)(x+3)}$ + … + $\frac{1}{(x+7)(x+8)}$ = $\frac{1}{14}$ Help 14/10/2021 Bởi Skylar $\frac{1}{(x+1)(x+2)}$ + $\frac{1}{(x+2)(x+3)}$ + … + $\frac{1}{(x+7)(x+8)}$ = $\frac{1}{14}$ Help
1/(x+1)(x+2) + … + 1/(x+7)(x+8) = 1/14 1/(x+1) – 1/(x+2) +1/(x+2) – 1/(x+3) +…+1/(x+7) – 1/(x+8) = 1/14 1/(x+1) – 1/(x+8) =1/14 7/(x+1)(x+8) = 1/14 x²+9x-90=0 x=6 hoặc x=-15 Bình luận
$\dfrac{1}{(x+1)(x+2)}+\dfrac{1}{(x+2)(x+3)}+…+\dfrac{1}{(x+7)(x+8)}=\dfrac{1}{14}$ ($x\ne -1;-2;…;-8$) $↔\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+…+\dfrac{1}{x+7}-\dfrac{1}{x+8}=\dfrac{1}{14}$ $↔\dfrac{1}{x+1}-\dfrac{1}{x+8}=\dfrac{1}{14}$ $↔\dfrac{x+8-x-1}{(x+1)(x+8)}=\dfrac{(x+1)(x+8)}{14(x+1)(x+8)}$ $↔98=x²+9x+8$ $↔0=x²+9x-90$ $↔0=(x-6)(x+15)$ $↔x-6=0\quad or\quad x+15=0$ $↔x=6(TM)\quad or\quad x=-15(TM)$ Bình luận
1/(x+1)(x+2) + … + 1/(x+7)(x+8) = 1/14
1/(x+1) – 1/(x+2) +1/(x+2) – 1/(x+3) +…+1/(x+7) – 1/(x+8) = 1/14
1/(x+1) – 1/(x+8) =1/14
7/(x+1)(x+8) = 1/14
x²+9x-90=0
x=6 hoặc x=-15
$\dfrac{1}{(x+1)(x+2)}+\dfrac{1}{(x+2)(x+3)}+…+\dfrac{1}{(x+7)(x+8)}=\dfrac{1}{14}$ ($x\ne -1;-2;…;-8$)
$↔\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+…+\dfrac{1}{x+7}-\dfrac{1}{x+8}=\dfrac{1}{14}$
$↔\dfrac{1}{x+1}-\dfrac{1}{x+8}=\dfrac{1}{14}$
$↔\dfrac{x+8-x-1}{(x+1)(x+8)}=\dfrac{(x+1)(x+8)}{14(x+1)(x+8)}$
$↔98=x²+9x+8$
$↔0=x²+9x-90$
$↔0=(x-6)(x+15)$
$↔x-6=0\quad or\quad x+15=0$
$↔x=6(TM)\quad or\quad x=-15(TM)$