toan nong hoi day

  • Thread starter menhlenhphuonghoang2312
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thienhoang_

$\leftrightarrow \dfrac{x+2}{327} +1+ \dfrac{x+3}{326} +1+ \dfrac{x+4}{325} +1+\dfrac{x+5}{324} +1+ \dfrac{x+349}{5} -4 = 0$

$\leftrightarrow \dfrac{x+ 329}{327} + \dfrac{x+ 329}{326} + \dfrac{x+ 329}{325}+ \dfrac{x+ 329}{324} + \dfrac{x+ 329}{5} =0$

$\leftrightarrow (x+ 329).(\dfrac{1}{327} + \dfrac{1}{326} + \dfrac{1}{325} + \dfrac{1}{324}+\dfrac{1}{5}) =0$

Do $( \dfrac{1}{327} +\dfrac{1}{326} + \dfrac{1}{325} + \dfrac{1}{324} +\dfrac{1}{5}) >0$ nên $x+ 329 =0 \rightarrow x= -329$
 
H

huongbloom

↔x+2/327+1+x+3/326+1+x+4/325+1+x+5/324+1+x+3/495−4=0

↔x+329/327+x+329/326+x+329/325+x+329/324+x+329/5=0

↔(x+329).(1/327+1/326+1/325+1/324+1/5)=0

Do (1/327+1/326+1/325+1/324+1/5)>0 nên x+329=0→x=−329
 
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