1. $\frac{x-6}{7}$ +$\frac{x-7}{8}$+ $\frac{x-8}{9}$=$\frac{x-9}{10}$+ $\frac{x-10}{11}$+ $\frac{x-11}{12}$
2. $\frac{x+32}{11}$+ $\frac{x+23}{12}$= $\frac{x+38}{13}$+ $\frac{x+27}{14}$
3.
1. $\frac{x-6}{7}$ +$\frac{x-7}{8}$+ $\frac{x-8}{9}$=$\frac{x-9}{10}$+ $\frac{x-10}{11}$+ $\frac{x-11}{12}$
2. $\frac{x+32}{11}$+ $\frac{x+23}{12}$= $\frac{x+38}{13}$+ $\frac{x+27}{14}$
3.
Đáp án:
Giải thích các bước giải:
`1)\frac{x-6}{7} +\frac{x-7}{8}+ \frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+ \frac{x-11}{12}`
`⇔\frac{x-6}{7} +\frac{x-7}{8}+ \frac{x-8}{9}+3=\frac{x-9}{10}+\frac{x-10}{11}+ \frac{x-11}{12}+3`
`⇔\frac{x-6}{7} +1 +\frac{x-7}{8}+1+ \frac{x-8}{9}+1=\frac{x-9}{10}+1+\frac{x-10}{11}+1+ \frac{x-11}{12}+1`
`⇔\frac{x-6}{7} +7/7 +\frac{x-7}{8}+8/8+ \frac{x-8}{9}+9/9=\frac{x-9}{10}+10/10+\frac{x-10}{11}+ 11/11+ \frac{x-11}{12}+ 12/12`
`⇔\frac{x+1}{7} +\frac{x+1}{8}+ \frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+ \frac{x+1}{12}`
`⇔\frac{x+1}{7} +\frac{x+1}{8}+ \frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}- \frac{x+1}{12}=0`
`⇔(x+1)(1/7+1/8+1/9-1/10-1/11-1/12)=0`
`⇔x+1=0` ( vì `1/7+1/8+1/9-1/10-1/11-1/12\ne0` )
`⇔x=0-1`
`⇔x=-1.`
Vậy `x=-1.`
`2) \frac{x+32}{11}+ \frac{x+23}{12}= \frac{x+38}{13}+ \frac{x+27}{14}`
`⇔\frac{x+32}{11}+ \frac{x+23}{12}-5= \frac{x+38}{13}+ \frac{x+27}{14}-5`
`⇔\frac{x+32}{11}-3+ \frac{x+23}{12}-2= \frac{x+38}{13}-3+ \frac{x+27}{14}-2`
`⇔\frac{x+32}{11}- 33/11+ \frac{x+23}{12}-24/12= \frac{x+38}{13}-39/13+ \frac{x+27}{14}-28/14`
`⇔\frac{x-1}{11}+ \frac{x-1}{12}= \frac{x-1}{13}+ \frac{x-1}{14}`
`⇔\frac{x-1}{11}+ \frac{x-1}{12}-\frac{x-1}{13}- \frac{x-1}{14}=0`
`⇔(x-1)(1/11+1/12-1/13-1/14)=0`
`⇔x-1=0` ( vì `1/11+1/12-1/13-1/14\ne0` )
`⇔x=0+1`
`⇔x=1.`
Vậy `x=1.`