Question

\(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\)

Answer 1

\(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\)

<=>\(\dfrac{x-342}{15}-1+\dfrac{x-323}{17}-2+\dfrac{x-300}{19}-3+\dfrac{x-273}{21}-4=0\)

<=>\(\dfrac{x-357}{15}+\dfrac{x-357}{17}+\dfrac{x-357}{19}+\dfrac{x-357}{21}=0\)

<=>\(\left(x-357\right)\left(\dfrac{1}{15}+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}\right)=0\)

vì 1/15+1/17+1/19+1/21 khác 0=>x-357=0<=>x=357

vậy..................

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