Question Number 159581 by mathlove last updated on 19/Nov/21
Commented by som(math1967) last updated on 19/Nov/21
![(9/3)=3 [when a,b,c≠0 ]](https://www.tinkutara.com/question/Q159582.png)
$$\frac{\mathrm{9}}{\mathrm{3}}=\mathrm{3}\:\:\left[{when}\:{a},{b},{c}\neq\mathrm{0}\:\right] \\ $$
Commented by mathlove last updated on 19/Nov/21
$${soltiuon}? \\ $$
Commented by tounghoungko last updated on 19/Nov/21
![a=b=c =k ⇒ ((k^2 +5k^2 +3k^2 )/(3k.k)) = (9/3)=3 [k≠ 0]](https://www.tinkutara.com/question/Q159584.png)
$${a}={b}={c}\:={k}\:\Rightarrow\:\frac{{k}^{\mathrm{2}} +\mathrm{5}{k}^{\mathrm{2}} +\mathrm{3}{k}^{\mathrm{2}} }{\mathrm{3}{k}.{k}}\:=\:\frac{\mathrm{9}}{\mathrm{3}}=\mathrm{3} \\ $$$$\:\left[{k}\neq\:\mathrm{0}\right]\: \\ $$
Answered by som(math1967) last updated on 19/Nov/21
![a^2 +b^2 +c^2 =ab+bc+ca ⇒2a^2 +2b^2 +2c^2 −2ab−2bc−2ca=0 ⇒(a^2 +b^2 −2ab)+(b^2 +c^2 −2bc)+(c^2 +a^2 −2ca)=0 ⇒(a−b)^2 +(b−c)^2 +(c−a)^2 =0 if a,b,c∈R then(a−b)^2 =0 (b−c)^2 =0,(c−a)^2 =0 ∴(a−b)=0⇒a=b,(b−c)=0⇒b=c ∴a=b=c now ((ab+5bc+3ac)/((a+b+c)(a+b−c))) =((a^2 +5a^2 +3a^2 )/((a+a+a)(a+a−a))) [ ∵a=b=c] =((9a^2 )/(3a^2 ))=3 [when a,b,c≠0]](https://www.tinkutara.com/question/Q159585.png)
$$\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} ={ab}+{bc}+{ca} \\ $$$$\Rightarrow\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} −\mathrm{2}{ab}−\mathrm{2}{bc}−\mathrm{2}{ca}=\mathrm{0} \\ $$$$\Rightarrow\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\right)+\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}\right)+\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}{ca}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({a}−{b}\right)^{\mathrm{2}} +\left({b}−{c}\right)^{\mathrm{2}} +\left({c}−{a}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${if}\:{a},{b},{c}\in{R}\:\:\:{then}\left({a}−{b}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({b}−{c}\right)^{\mathrm{2}} =\mathrm{0},\left({c}−{a}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\therefore\left({a}−{b}\right)=\mathrm{0}\Rightarrow{a}={b},\left({b}−{c}\right)=\mathrm{0}\Rightarrow{b}={c} \\ $$$$\therefore{a}={b}={c} \\ $$$$\:{now}\:\frac{{ab}+\mathrm{5}{bc}+\mathrm{3}{ac}}{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)} \\ $$$$=\frac{{a}^{\mathrm{2}} +\mathrm{5}{a}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} }{\left({a}+{a}+{a}\right)\left({a}+{a}−{a}\right)}\:\:\:\left[\:\because{a}={b}={c}\right] \\ $$$$=\frac{\mathrm{9}{a}^{\mathrm{2}} }{\mathrm{3}{a}^{\mathrm{2}} }=\mathrm{3}\:\:\:\:\left[{when}\:{a},{b},{c}\neq\mathrm{0}\right] \\ $$