$\frac{x+1}{10}$ + $\frac{x+1}{11}$ + $\frac{x+1}{12}$ = 0 Giúp mình với 27/08/2021 Bởi Josephine $\frac{x+1}{10}$ + $\frac{x+1}{11}$ + $\frac{x+1}{12}$ = 0 Giúp mình với
$\frac{x+1}{10}$ + $\frac{x+1}{11}$ + $\frac{x+1}{12}$ =0 <=> (x+1) ($\frac{1}{10}$ + $\frac{1}{11}$ + $\frac{1}{12}$ =0 <=> (x+1) $\frac{181}{660}$ = 0 => x+1=0 => x=-1 Vote 5s và ctlhn choa mị nhé!!! ^ω^ Bình luận
$\dfrac{x+1}{10} + \dfrac{x +1}{11} + \dfrac{x+1}{12} = 0$ ⇔$(x+1)\bigg(\dfrac{1}{10} + \dfrac{1}{11} + \dfrac{1}{12}\bigg) = 0$ ⇔$(x+1) = 0$ (Vì $\dfrac{1}{10} + \dfrac{1}{11} + \dfrac{1}{12} > 0$) ⇔$ x = -1$ Bình luận
$\frac{x+1}{10}$ + $\frac{x+1}{11}$ + $\frac{x+1}{12}$ =0
<=> (x+1) ($\frac{1}{10}$ + $\frac{1}{11}$ + $\frac{1}{12}$ =0
<=> (x+1) $\frac{181}{660}$ = 0
=> x+1=0 => x=-1
Vote 5s và ctlhn choa mị nhé!!! ^ω^
$\dfrac{x+1}{10} + \dfrac{x +1}{11} + \dfrac{x+1}{12} = 0$
⇔$(x+1)\bigg(\dfrac{1}{10} + \dfrac{1}{11} + \dfrac{1}{12}\bigg) = 0$
⇔$(x+1) = 0$ (Vì $\dfrac{1}{10} + \dfrac{1}{11} + \dfrac{1}{12} > 0$)
⇔$ x = -1$