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SaNNy322
@SaNNy322
August 2022
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Решите Пожалуйста (Фото)
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selenadior
[tex]1- \frac{2b+1}{b^{2}+1 }= \frac{(b^{2}+1)*(b+1)-(b+1)*(2b+1)-b*(b^{2}+1) }{(b^{2}+1)*(b+1) }= \frac{b^{3}+b^{2}+b+1-(2b^{2}+3b+1)-b^3-b}{(b^{2}+1)*(b+1) }= \frac{b^3+b^2+b+1-(2b^2+3b+1)-b^3-b}{(b^2+1)*(b+1)}= \frac{b^2+1-2b^2-3b-1}{(b^2+1)*(b+1)}= \frac{-b^2-3b-1}{(b^2+1)*(b+1} /tex]
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