Đáp án:
$\frac{x+1}{39}$ + $\frac{x+2}{38}$+$\frac{x+3}{37}$+ $\frac{x-9}{49}$+ $\frac{x-8}{48}$ = -5
⇔($\frac{x+1}{39}$+1) + ($\frac{x+2}{38}$+1) + ($\frac{x+3}{37}$+1)+ ($\frac{x-9}{49}$+1)+ ($\frac{x-8}{48}$+1) = 0
⇔$\frac{x+1+39}{39}$ + $\frac{x+2+38}{38}$+$\frac{x+3+37}{37}$+ $\frac{x-9+49}{49}$+ $\frac{x-8+48}{48}$ = 0
⇔$\frac{x+40}{39}$ + $\frac{x+40}{38}$+$\frac{x+40}{37}$+ $\frac{x+40}{49}$+ $\frac{x+40}{48}$ = 0
⇔ ( x+40) ( $\frac{1}{39}$ + $\frac{1}{38}$+$\frac{1}{37}$+ $\frac{1}{49}$+ $\frac{1}{48}$) = 0
⇒ x+ 40 = 0 ( vì $\frac{1}{39}$ + $\frac{1}{38}$+$\frac{1}{37}$+ $\frac{1}{49}$+ $\frac{1}{48}$ >0)
⇒ x =-40
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