a) $\frac{x+1}{10}$ + $\frac{x+1}{11}$ + $\frac{x+1}{12}$ = $\frac{x+1}{13}$ + $\frac{x+1}{14}$ 31/07/2021 Bởi Autumn a) $\frac{x+1}{10}$ + $\frac{x+1}{11}$ + $\frac{x+1}{12}$ = $\frac{x+1}{13}$ + $\frac{x+1}{14}$
$\frac{x+1}{10}$ + $\frac{x+1}{11}$ + $\frac{x+1}{12}$ = $\frac{x+1}{13}$ + $\frac{x+1}{14}$ => $\frac{x+1}{10}$ + $\frac{x+1}{11}$ + $\frac{x+1}{12}$ – $\frac{x+1}{13}$ – $\frac{x+1}{14}$ =0 ⇒(x+1)( $\frac{1}{10}$ + $\frac{1}{11}$ + $\frac{1}{12}$ – $\frac{1}{13}$ – $\frac{1}{14}$ )=0 vì $\frac{1}{10}$ + $\frac{1}{11}$ + $\frac{1}{12}$ – $\frac{1}{13}$ – $\frac{1}{14}$ >0 nên x+1 =0 =>x=-1 @htkbaam Bình luận
$\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}$ $\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0$ $(x+1).(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14})=0$ Vì $\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}>0$ $→x+1=0$ $→x=0-1=-1$ Vậy $x=-1$ Bình luận
$\frac{x+1}{10}$ + $\frac{x+1}{11}$ + $\frac{x+1}{12}$ = $\frac{x+1}{13}$ + $\frac{x+1}{14}$
=> $\frac{x+1}{10}$ + $\frac{x+1}{11}$ + $\frac{x+1}{12}$ – $\frac{x+1}{13}$ – $\frac{x+1}{14}$ =0
⇒(x+1)( $\frac{1}{10}$ + $\frac{1}{11}$ + $\frac{1}{12}$ – $\frac{1}{13}$ – $\frac{1}{14}$ )=0
vì $\frac{1}{10}$ + $\frac{1}{11}$ + $\frac{1}{12}$ – $\frac{1}{13}$ – $\frac{1}{14}$ >0
nên x+1 =0
=>x=-1
@htkbaam
$\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}$
$\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0$
$(x+1).(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14})=0$
Vì $\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}>0$
$→x+1=0$
$→x=0-1=-1$
Vậy $x=-1$