Tìm x biết: $\frac{55-x}{1963}$ + $\frac{50-x}{1968}$ + $\frac{45-x}{1973}$ + $\frac{40-x}{1978}$ + 4 = 0

0 bình luận về “Tìm x biết: $\frac{55-x}{1963}$ + $\frac{50-x}{1968}$ + $\frac{45-x}{1973}$ + $\frac{40-x}{1978}$ + 4 = 0”

1. $\dfrac{55-x}{1963}+\dfrac{50-x}{1968}+\dfrac{45-x}{1973}+\dfrac{40-x}{1978}+4=0 \\⇔\dfrac{55-x}{1963}+1+\dfrac{50-x}{1968}+1+\dfrac{45-x}{1973}+1+\dfrac{40-x}{1978}+1=0 \\⇔\dfrac{55-x+1963}{1963}+\dfrac{50-x+1968}{1968}+\dfrac{45-x+1973}{1973}+\dfrac{40-x}{1978}=0 \\⇔\dfrac{2018-x}{1963}+\dfrac{2018-x}{1968}+\dfrac{2018-x}{1973}+\dfrac{2018-x}{1978}=0 \\⇔(2018-x)\bigg(\dfrac{1}{1963}+\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}\bigg)=0 \\⇒2018-x=0 \\⇔x=2018 \\Vậy\ x=2018$

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2. Em tham khảo

   $\dfrac{55-x}{1963}+\dfrac{50-x}{1968}+\dfrac{45-x}{1973}+\dfrac{40-x}{1978}+4=0$

   $⇒\dfrac{55-x}{1963}+\dfrac{50-x}{1968}+\dfrac{45-x}{1973}+\dfrac{40-x}{1978}+1+1+1+1=0$

   $⇒(\dfrac{55-x}{1963}+1)+(\dfrac{50-x}{1968}+1)+(\dfrac{45-x}{1973}+1)+(\dfrac{40-x}{1978}+1)=0$

   $⇒\dfrac{55-x+1963}{1963}+\dfrac{50-x+1968}{1968}+\dfrac{45-x+1973}{1973}+\dfrac{40-x+1978}{1978}=0$

   $⇒\dfrac{2018-x}{1963}+\dfrac{2018-x}{1968}+\dfrac{2018-x}{1973}+\dfrac{2018-x}{1978}=0$

   $⇒(2018-x)×(\dfrac{1}{1963}+\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978})$

   $⇔(2018-x)=0$

   $⇒x=2018$

   Để $0×(\dfrac{1}{1963}+\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978})=0$

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