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Question-205380

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Question Number 205380 by cortano12 last updated on 19/Mar/24
Commented by Rasheed.Sindhi last updated on 19/Mar/24
please don′t write your solution as a comment! Instead write it as answer in order to prserve chronological order of answers!
$$\boldsymbol{{please}}\:{don}'{t}\:{write}\:{your}\:{solution} \\ $$$${as}\:{a}\:{comment}!\:{Instead}\:{write}\:{it}\:{as} \\ $$$${answer}\:\boldsymbol{{in}}\:\boldsymbol{{order}}\:\boldsymbol{{to}}\:\boldsymbol{{prserve}}\:\boldsymbol{{chronological}} \\ $$$$\boldsymbol{{order}}\:\boldsymbol{{of}}\:\boldsymbol{{answers}}! \\ $$
Answered by A5T last updated on 19/Mar/24
((Σ(1+bc+b)(1+ca+c)=Σ(2+ca+2c+2bc+bc^2 +b))/((2+ca+2c+2bc+bc^2 +b)(1+ab+a))) =((6+3(ca+ab+bc)+3(a+b+c)+bc^2 +ca^2 +ab^2 )/(6+3ca+3c+3bc+bc^2 +3b+3ab+3a+a^2 c+ab^2 )) =1
$$\frac{\Sigma\left(\mathrm{1}+{bc}+{b}\right)\left(\mathrm{1}+{ca}+{c}\right)=\Sigma\left(\mathrm{2}+{ca}+\mathrm{2}{c}+\mathrm{2}{bc}+{bc}^{\mathrm{2}} +{b}\right)}{\left(\mathrm{2}+{ca}+\mathrm{2}{c}+\mathrm{2}{bc}+{bc}^{\mathrm{2}} +{b}\right)\left(\mathrm{1}+{ab}+{a}\right)} \\ $$$$=\frac{\mathrm{6}+\mathrm{3}\left({ca}+{ab}+{bc}\right)+\mathrm{3}\left({a}+{b}+{c}\right)+{bc}^{\mathrm{2}} +{ca}^{\mathrm{2}} +{ab}^{\mathrm{2}} }{\mathrm{6}+\mathrm{3}{ca}+\mathrm{3}{c}+\mathrm{3}{bc}+{bc}^{\mathrm{2}} +\mathrm{3}{b}+\mathrm{3}{ab}+\mathrm{3}{a}+{a}^{\mathrm{2}} {c}+{ab}^{\mathrm{2}} } \\ $$$$=\mathrm{1} \\ $$
Answered by Rasheed.Sindhi last updated on 19/Mar/24
abc=1 (1/(1+ab+a))+(1/(1+bc+b))+(1/(1+ca+c)) =((1(c))/((1+ab+a)(c)))+((1(ca))/((1+bc+b)(ca)))+((1×1)/((1+ca+c)(1))) =(c/(c+abc+ca))+((ca)/(ca+abc.c+abc))+(1/(1+ca+c)) =(c/(c+1+ca))+((ca)/(ca+c+1))+(1/(1+ca+c)) =((c+ca+1)/(c+ca+1))=1
$${abc}=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{ab}+{a}}+\frac{\mathrm{1}}{\mathrm{1}+{bc}+{b}}+\frac{\mathrm{1}}{\mathrm{1}+{ca}+{c}} \\ $$$$=\frac{\mathrm{1}\left({c}\right)}{\left(\mathrm{1}+{ab}+{a}\right)\left({c}\right)}+\frac{\mathrm{1}\left({ca}\right)}{\left(\mathrm{1}+{bc}+{b}\right)\left({ca}\right)}+\frac{\mathrm{1}×\mathrm{1}}{\left(\mathrm{1}+{ca}+{c}\right)\left(\mathrm{1}\right)} \\ $$$$=\frac{{c}}{{c}+{abc}+{ca}}+\frac{{ca}}{{ca}+{abc}.{c}+{abc}}+\frac{\mathrm{1}}{\mathrm{1}+{ca}+{c}} \\ $$$$=\frac{{c}}{{c}+\mathrm{1}+{ca}}+\frac{{ca}}{{ca}+{c}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{1}+{ca}+{c}} \\ $$$$=\frac{{c}+{ca}+\mathrm{1}}{{c}+{ca}+\mathrm{1}}=\mathrm{1} \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 19/Mar/24
abc=1⇒ { ((bc=(1/a))),((c=(1/(ab)))) :} (1/(1+ab+a))+(1/(1+bc+b))+(1/(1+ca+c)) =(1/(1+ab+a))+(1/(1+(1/a)+b))+(1/(1+ca+(1/(ab)))) (1/(1+ab+a))+(a/(a+1+ab))+((ab)/(ab+a.abc+1)) =((1+a+ab)/(1+a+ab))=1
$${abc}=\mathrm{1}\Rightarrow\begin{cases}{{bc}=\frac{\mathrm{1}}{{a}}}\\{{c}=\frac{\mathrm{1}}{{ab}}}\end{cases} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{ab}+{a}}+\frac{\mathrm{1}}{\mathrm{1}+{bc}+{b}}+\frac{\mathrm{1}}{\mathrm{1}+{ca}+{c}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+{ab}+{a}}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{a}}+{b}}+\frac{\mathrm{1}}{\mathrm{1}+{ca}+\frac{\mathrm{1}}{{ab}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{ab}+{a}}+\frac{{a}}{{a}+\mathrm{1}+{ab}}+\frac{{ab}}{{ab}+{a}.{abc}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}+{a}+{ab}}{\mathrm{1}+{a}+{ab}}=\mathrm{1} \\ $$

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