a ) $\frac{x + 2}{11}$ + $\frac{x + 2}{12}$ + $\frac{x + 2}{13}$ = $\frac{x + 2}{14}$ + $\frac{x + 2}{15}$ b ) $\frac{x + 3}{35}$ + $\frac{x + 7}{

a ) $\frac{x + 2}{11}$ + $\frac{x + 2}{12}$ + $\frac{x + 2}{13}$ = $\frac{x + 2}{14}$ + $\frac{x + 2}{15}$
b ) $\frac{x + 3}{35}$ + $\frac{x + 7}{31}$ + $\frac{x + 11}{27}$ = $\frac{x + 22}{16}$ + $\frac{x – 27}{65}$ + $\frac{x + 1}{37}$
c ) $\frac{x + 2}{327}$ + $\frac{x + 3}{326}$ + $\frac{x + 4}{325}$ + $\frac{x + 5}{324}$ +$\frac{x + 34}{5}$ = 0

0 bình luận về “a ) $\frac{x + 2}{11}$ + $\frac{x + 2}{12}$ + $\frac{x + 2}{13}$ = $\frac{x + 2}{14}$ + $\frac{x + 2}{15}$ b ) $\frac{x + 3}{35}$ + $\frac{x + 7}{”

  1. a/ $\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}=\dfrac{x+2}{14}+\dfrac{x+2}{15}$
    ⇔ $\dfrac{x+2}{11}+\dfrac{x+2}{12}+\dfrac{x+2}{13}-\dfrac{x+2}{14}-\dfrac{x+2}{15}=0$
    ⇔ $(x+2)(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15})=0$
    Vì $\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15} > 0$
    nên $x+2=0$
    ⇔ $x=-2$

    b/ $\dfrac{x+3}{35}+\dfrac{x+7}{31}+\dfrac{x+11}{27}=\dfrac{x+22}{16}+\dfrac{x-27}{65}+\dfrac{x+1}{37}$
    ⇔ $\dfrac{x+3}{35}+1+\dfrac{x+7}{31}+1+\dfrac{x+11}{27}+1-\dfrac{x+22}{16}-1-\dfrac{x-27}{65}-1-\dfrac{x+1}{37}-1=0$
    ⇔ $\dfrac{x+38}{35}+\dfrac{x+38}{31}+\dfrac{x+38}{27}-…….-\dfrac{x+38}{27}=0$
    ⇔ $(x+38)(\dfrac{1}{35}+\dfrac{1}{31}+…..-\dfrac{1}{27})=0$
    Vì $\dfrac{1}{35}+\dfrac{1}{31}+…..-\dfrac{1}{27} \neq 0$
    nên $x+38=0$
    ⇔ $x=-38$

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