giải phương trình x+1/2008+x+2/2007=x+3/2006+x+4/2005
2 câu trả lời
Giải thích các bước giải:
`\frac{x+1}{2008}+\frac{x+2}{2007}=\frac{x+3}{2006}+\frac{x+4}{2005}`
<=> `(1+\frac{x+1}{2008})+(1+\frac{x+2}{2007})=(1+\frac{x+3}{2006})+(1+\frac{x+4}{2005})`
<=> `\frac{x+2009}{2008}+\frac{x+2009}{2007}=\frac{x+2009}{2006}+\frac{x+2009}{2005}`
<=> `\frac{x+2009}{2008}+\frac{x+2009}{2007}-\frac{x+2009}{2006}-\frac{x+2009}{2005}=0`
<=> `(x+2009)(\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2006}-\frac{1}{2005})=0`
mà `\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2006}-\frac{1}{2005}≠0`
=> `x+2009=0`
Vậy `x=-2009`
@Deawoo
`\frac{x+1}{2008}+\frac{x+2}{2007}=\frac{x+3}{2006}+\frac{x+4}{2005}`
⇔`\frac{x+1}{2008}+1+\frac{x+2}{2007}+1=\frac{x+3}{2006}+1+\frac{x+4}{2005}+1`
⇔`\frac{x+2009}{2008}+\frac{x+2009}{2007}=\frac{x+2009}{2006}+\frac{x+2009}{2005}`
⇔`(x+2009)(1/2008+1/2009-1/2006-1/2005)=0`
Do `1/2008+1/2009-1/2006-1/2005 \ne 0` nên `x+2009=0`
⇔`x=-2009`
Vậy `x=-2009`