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Question Number 161126 by geron last updated on 12/Dec/21

Commented by geron last updated on 12/Dec/21

79 ?

$$\mathrm{79}\:? \\ $$

Commented by geron last updated on 12/Dec/21

P(x):Q(x)=f(x)+k, k=?

$$\boldsymbol{{P}}\left({x}\right):\boldsymbol{{Q}}\left({x}\right)={f}\left({x}\right)+\boldsymbol{{k}},\:\boldsymbol{{k}}=? \\ $$

Answered by TheSupreme last updated on 12/Dec/21

79 (x^2 +2014x+2015)(x^2 +2015x+2016)+(x^2 +2013x+2014)(x^2 +2017x+2018)+3x+2/(x^2 +2016x+2017) (P(x)−2x−2)(P(x)−x−1)+(P(x)−3x−3)(P(x)+x+1)+3x+2 (−2x−2)(−x−1)+(−3x−3)(x+1)+3x+2= =2(x+1)^2 −3(x+1)^2 +3x+2= =−x^2 −2x−1+3x+2=−x^2 +x+1/(x^2 +2016x+2017) =2017x+2018

$$\mathrm{79} \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{2014}{x}+\mathrm{2015}\right)\left({x}^{\mathrm{2}} +\mathrm{2015}{x}+\mathrm{2016}\right)+\left({x}^{\mathrm{2}} +\mathrm{2013}{x}+\mathrm{2014}\right)\left({x}^{\mathrm{2}} +\mathrm{2017}{x}+\mathrm{2018}\right)+\mathrm{3}{x}+\mathrm{2}/\left({x}^{\mathrm{2}} +\mathrm{2016}{x}+\mathrm{2017}\right) \\ $$$$\left({P}\left({x}\right)−\mathrm{2}{x}−\mathrm{2}\right)\left({P}\left({x}\right)−{x}−\mathrm{1}\right)+\left({P}\left({x}\right)−\mathrm{3}{x}−\mathrm{3}\right)\left({P}\left({x}\right)+{x}+\mathrm{1}\right)+\mathrm{3}{x}+\mathrm{2} \\ $$$$\left(−\mathrm{2}{x}−\mathrm{2}\right)\left(−{x}−\mathrm{1}\right)+\left(−\mathrm{3}{x}−\mathrm{3}\right)\left({x}+\mathrm{1}\right)+\mathrm{3}{x}+\mathrm{2}= \\ $$$$=\mathrm{2}\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}= \\ $$$$=−{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}+\mathrm{3}{x}+\mathrm{2}=−{x}^{\mathrm{2}} +{x}+\mathrm{1}/\left({x}^{\mathrm{2}} +\mathrm{2016}{x}+\mathrm{2017}\right) \\ $$$$=\mathrm{2017}{x}+\mathrm{2018} \\ $$

Commented by Rasheed.Sindhi last updated on 13/Dec/21

Typo sir: In 3rd line you′ve written P(x) instead of Q(x).

$$\mathrm{Typo}\:\mathrm{sir}:\:{In}\:\mathrm{3}{rd}\:{line}\:{you}'{ve}\:{written} \\ $$$${P}\left({x}\right)\:{instead}\:{of}\:{Q}\left({x}\right). \\ $$

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