Tìm x: $\frac{x-1}{8}$ + $\frac{x-1}{9}$ = $\frac{x-1}{10}$ +$\frac{x-1}{11}$ 25/07/2021 Bởi Rose Tìm x: $\frac{x-1}{8}$ + $\frac{x-1}{9}$ = $\frac{x-1}{10}$ +$\frac{x-1}{11}$
Em tham khảo: $\frac{x-1}{8}+$$\frac{x-1}{9}=$$\frac{x-1}{10}+$$\frac{x-1}{11}$ ⇔ $\frac{x-1}{8}+$$\frac{x-1}{9}-$$\frac{x-1}{10}-$$\frac{x-1}{11}=0$ ⇔$(x-1)$.($\frac{1}{8}+$$\frac{1}{9}-$$\frac{1}{10}-$$\frac{1}{11}$) $=0$ Vì $\frac{1}{8}+$$\frac{1}{9}-$$\frac{1}{10}-$$\frac{1}{11}>0$ ⇔$x-1=0$ ⇔$x=1$ HỌC TỐT @Chin……….. Bình luận
$\dfrac{x-1}{8}+\dfrac{x-1}{9}=\dfrac{x-1}{10}+\dfrac{x-1}{11}$ $\dfrac{x-1}{8}+\dfrac{x-1}{9}-\dfrac{x-1}{10}-\dfrac{x-1}{11}=0$ $(x-1).(\dfrac{1}{8}+\dfrac{1}{9}-\dfrac{1}{10}-\dfrac{1}{11})=0$ Vì $\dfrac{1}{8}+\dfrac{1}{9}-\dfrac{1}{10}-\dfrac{1}{11}>0$ $→x-1=0$ $→x=1$ Vậy $x=1$ Bình luận
Em tham khảo:
$\frac{x-1}{8}+$$\frac{x-1}{9}=$$\frac{x-1}{10}+$$\frac{x-1}{11}$
⇔ $\frac{x-1}{8}+$$\frac{x-1}{9}-$$\frac{x-1}{10}-$$\frac{x-1}{11}=0$
⇔$(x-1)$.($\frac{1}{8}+$$\frac{1}{9}-$$\frac{1}{10}-$$\frac{1}{11}$) $=0$
Vì $\frac{1}{8}+$$\frac{1}{9}-$$\frac{1}{10}-$$\frac{1}{11}>0$
⇔$x-1=0$
⇔$x=1$
HỌC TỐT
@Chin………..
$\dfrac{x-1}{8}+\dfrac{x-1}{9}=\dfrac{x-1}{10}+\dfrac{x-1}{11}$
$\dfrac{x-1}{8}+\dfrac{x-1}{9}-\dfrac{x-1}{10}-\dfrac{x-1}{11}=0$
$(x-1).(\dfrac{1}{8}+\dfrac{1}{9}-\dfrac{1}{10}-\dfrac{1}{11})=0$
Vì $\dfrac{1}{8}+\dfrac{1}{9}-\dfrac{1}{10}-\dfrac{1}{11}>0$
$→x-1=0$
$→x=1$
Vậy $x=1$