Menu Close

Question-150870




Question Number 150870 by liberty last updated on 16/Aug/21
Commented by mnjuly1970 last updated on 16/Aug/21
  pls  solve it...
$$\:\:{pls}\:\:{solve}\:{it}… \\ $$
Answered by tabata last updated on 16/Aug/21
if the equation is given by :    x^3 −2x^2 −x+2=0⇒(x+1)(x−1)(x−2)=0    ⇒x_1 =−1=a , x_2 =1=b , x_3 =2=c    ∴(1/(a(b^2 +c^2 −a^2 )))+(1/(b(c^2 +a^2 −b^2 )))+(1/(c(a^2 +b^2 −c^2 )))    =−(1/4)+(1/4)−(1/4)=−(1/4)    ⟨M.T ⟩
$${if}\:{the}\:{equation}\:{is}\:{given}\:{by}\:: \\ $$$$ \\ $$$${x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −{x}+\mathrm{2}=\mathrm{0}\Rightarrow\left({x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)=\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow{x}_{\mathrm{1}} =−\mathrm{1}={a}\:,\:{x}_{\mathrm{2}} =\mathrm{1}={b}\:,\:{x}_{\mathrm{3}} =\mathrm{2}={c} \\ $$$$ \\ $$$$\therefore\frac{\mathrm{1}}{{a}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}+\frac{\mathrm{1}}{{b}\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}+\frac{\mathrm{1}}{{c}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)} \\ $$$$ \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$$$\langle{M}.{T}\:\rangle \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *