Question

[le thi quynhanh]

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

Answer 1

[Xyz OLM]

Ta có : \(\frac{x-342}{15}+\frac{x-323}{17}+\frac{x-300}{19}+\frac{x-273}{21}=10\)

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

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

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

Vì \(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\ne0\)

=> x - 357 = 0

=> x = 357

Vậy x = 357