Đáp án + Giải thích các bước giải:
`a,\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}`
`⇔(\frac{x+1}{35}+1)+(\frac{x+3}{33}+1)=(\frac{x+5}{31}+1)+(\frac{x+7}{29}+1)`
`⇔(\frac{x+1}{35}+\frac{35}{35})+(\frac{x+3}{33}+\frac{33}{33})=(\frac{x+5}{31}+\frac{31}{31)+(\frac{x+7}{29}+\frac{29}{29})`
`⇔\frac{x+36}{35}+\frac{x+36}{33}=\frac{x+36}{31}+\frac{x+36}{29}`
`⇔\frac{x+36}{35}+\frac{x+36}{33}-\frac{x+36}{31}-\frac{x+36}{29}=0`
`⇔(x+36)(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29})=0`
`⇔x+36=0` `[ Do (\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29})<0 ]`
`⇔x=-36`
Vậy phương trình có một nghiệm duy nhất là : `x=-36`
`----------------`
`b,\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}`
`⇔(\frac{x+1}{65}+1)+(\frac{x+3}{63}+1)=(\frac{x+5}{61}+1)+(\frac{x+7}{59}+1)`
`⇔(\frac{x+1}{65}+\frac{65}{65})+(\frac{x+3}{63}+\frac{63}{63})=(\frac{x+5}{61}+\frac{61}{61})+(\frac{x+7}{59}+\frac{59}{59})`
`⇔\frac{x+66}{65}+\frac{x+66}{63}=\frac{x+66}{61}+\frac{x+66}{59}`
`⇔\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0`
`⇔(x+66)(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59})=0`
`⇔x+66=0` `[ Do (\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59})<0 ]`
`⇔x=-66`
Vậy phương trình có một nghiệm duy nhất là : `x=-66`
`----------------`
`c,\frac{x-2021}{2009}+\frac{x-2022}{2008}+\frac{x-2023}{2007}=\frac{x-2024}{2006}+\frac{x-2025}{2005}+\frac{x-2026}{2004}`
`⇔(\frac{x-2021}{2009}-1)+(\frac{x-2022}{2008}-1)+(\frac{x-2023}{2007}-1)=(\frac{x-2024}{2006}-1)+(\frac{x-2025}{2005}-1)+(\frac{x-2026}{2004}-1)`
`⇔(\frac{x-2021}{2009}-\frac{2009}{2009})+(\frac{x-2022}{2008}-\frac{2008}{2008})+(\frac{x-2023}{2007}-\frac{2007}{2007})=(\frac{x-2024}{2006}-\frac{2006}{2006})+(\frac{x-2025}{2005}-\frac{2005}{2005})+(\frac{x-2026}{2004}-\frac{2004}{2004})`
`⇔\frac{x-4030}{2009}+\frac{x-4030}{2008}+\frac{x-4030}{2007}=\frac{x-4030}{2006}+\frac{x-4030}{2005}+\frac{x-4030}{2004}`
`⇔\frac{x-4030}{2009}+\frac{x-4030}{2008}+\frac{x-4030}{2007}-\frac{x-4030}{2006}-\frac{x-4030}{2005}-\frac{x-4030}{2004}=0`
`⇔(x-4030)(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004})=0`
`⇔x-4030=0` `[ Do (\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004})<0 ]`
`⇔x=4030`
Vậy phương trình có một nghiệm duy nhất là : `x=4030`
