a, $\frac{x+3}{10}$ +$\frac{x+3}{11}$+ $\frac{x+3}{12}$ =$\frac{x+3}{13}$+ $\frac{x+3}{14}$
⇔$\frac{x+3}{10}$ +$\frac{x+3}{11}$+ $\frac{x+3}{12}$-$\frac{x+3}{13}$-$\frac{x+3}{14}$ = 0
⇔(x+3)($\frac{1}{10}$ +$\frac{1}{11}$ +$\frac{1}{12}$ -$\frac{1}{13}$ -$\frac{1}{14}$ )=0
⇔ x+3= 0
(Vì $\frac{1}{10}$ +$\frac{1}{11}$ +$\frac{1}{12}$ -$\frac{1}{13}$ -$\frac{1}{14}$$\neq$0)
⇔ x=-3
Vậy x = -3
b,$\frac{x-20}{9}$ +$\frac{x-21}{10}$ +$\frac{x-26}{15}$ =-3
⇔$\frac{10(x-20)}{90}$ +$\frac{9(x-21)}{90}$ +$\frac{6(x-26)}{90}$ =-3
⇔$\frac{10(x-20)+9(x-21)+6(x-26)}{90}$ =-3
⇔$\frac{10x-200+9x-189+6x-156}{90}$=-3
⇔$\frac{(10x+9x+6x)-(200+189+156)}{90}$=-3
⇔$\frac{25x-545}{90}$ =-3
⇔ 25x-545 = -270
⇔ 25x = 275
⇔ x= 11
Vậy x = 11
c, $\frac{x+2014}{4}$ +$\frac{x+2013}{5}$+ $\frac{x+2012}{6}$= $\frac{x+2010}{8}$ +$\frac{x+2009}{9}$+ $\frac{x+2008}{10}$
⇔$\frac{x+2014}{4}$ +1+$\frac{x+2013}{5}$+1+ $\frac{x+2012}{6}$+1= $\frac{x+2010}{8}$ +1+$\frac{x+2009}{9}$+1+ $\frac{x+2008}{10}$ +1
⇔$\frac{x+2018}{4}$ +$\frac{x+2018}{5}$+ $\frac{x+2018}{6}$= $\frac{x+2018}{8}$ +$\frac{x+2018}{9}$+ $\frac{x+2018}{10}$
⇔$\frac{x+2018}{4}$ +$\frac{x+2018}{5}$+ $\frac{x+2018}{6}$- $\frac{x+2018}{8}$ -$\frac{x+2018}{9}$- $\frac{x+2018}{10}$ =0
⇔(x+2018)($\frac{1}{4}$+ $\frac{1}{5}$ +$\frac{1}{6}$- $\frac{1}{8}$- $\frac{1}{9}$- $\frac{1}{10}$ )= 0
⇔x+2018=0
(Vì $\frac{1}{4}$+ $\frac{1}{5}$ +$\frac{1}{6}$- $\frac{1}{8}$- $\frac{1}{9}$- $\frac{1}{10}$ $\neq$ 0
⇔ x = -2018
Vậy x = -2018
d, $\frac{x-2}{27}$+ $\frac{x-3}{26}$+ $\frac{x-4}{25}$+ $\frac{x-44}{5}$=0
⇔($\frac{x-2}{27}$-1)+( $\frac{x-3}{26}$-1)+( $\frac{x-4}{25}$-1)+ ($\frac{x-44}{5}$+3)=0
⇔$\frac{x-29}{27}$ + $\frac{x-29}{26}$ + $\frac{x-29}{25}$ + $\frac{x-29}{5}$=0
⇔(x-29)($\frac{1}{27}$+ $\frac{1}{26}$+ $\frac{1}{25}$+ $\frac{1}{5}$)=0
⇔x - 29 = 0
(Vì $\frac{1}{27}$+ $\frac{1}{26}$+ $\frac{1}{25}$+ $\frac{1}{5}$$\neq$ 0
⇔ x = 29
Vậy x = 29
Bạn cho mik xin ctlhn nhé!~~
