Question-174074

Question Number 174074 by AgniMath last updated on 24/Jul/22

Commented by infinityaction last updated on 24/Jul/22

ay = bx , cy =bz , cx = az ((ax−by)/((a+b)(x−y)))+((by−cz)/((b+c)(y−z)))+((cz−ax)/((c+a)(z−x))) ((ax−by)/((a+b)(x−y)))−1+((by−cz)/((b+c)(y−z)))−1+((cz−ax)/((c+a)(z−x)))−1+3 ((ay−bx)/((a+b)(x−y)))+((bz−cy)/((b+c)(y−z)))+((cx−az)/((c+a)(z−x)))+3 0+0+0+3 = 3

$${ay}\:=\:{bx}\:,\:{cy}\:={bz}\:,\:{cx}\:=\:{az} \\ $$$$\frac{{ax}−{by}}{\left({a}+{b}\right)\left({x}−{y}\right)}+\frac{{by}−{cz}}{\left({b}+{c}\right)\left({y}−{z}\right)}+\frac{{cz}−{ax}}{\left({c}+{a}\right)\left({z}−{x}\right)}\: \\ $$$$\frac{{ax}−{by}}{\left({a}+{b}\right)\left({x}−{y}\right)}−\mathrm{1}+\frac{{by}−{cz}}{\left({b}+{c}\right)\left({y}−{z}\right)}−\mathrm{1}+\frac{{cz}−{ax}}{\left({c}+{a}\right)\left({z}−{x}\right)}−\mathrm{1}+\mathrm{3} \\ $$$$\frac{{ay}−{bx}}{\left({a}+{b}\right)\left({x}−{y}\right)}+\frac{{bz}−{cy}}{\left({b}+{c}\right)\left({y}−{z}\right)}+\frac{{cx}−{az}}{\left({c}+{a}\right)\left({z}−{x}\right)}+\mathrm{3} \\ $$$$\:\mathrm{0}+\mathrm{0}+\mathrm{0}+\mathrm{3}\:=\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\: \\ $$

Answered by som(math1967) last updated on 24/Jul/22

let (x/a)=(y/b)=(z/c)=k [k≠0] ⇒x=ak,y=bk ,z=ck ((k(a^2 −b^2 ))/(k(a+b)(a−b))) +((k(b^2 −c^2 ))/(k(b+c)(b−c))) +((k(c^2 −a^2 ))/(k(c+a)(c−a))) =(((a^2 −b^2 ))/((a^2 −b^2 ))) +(((b^2 −c^2 ))/((b^2 −c^2 ))) +(((c^2 −a^2 ))/((c^2 −a^2 ))) =1+1+1=3

$${let}\:\frac{{x}}{{a}}=\frac{{y}}{{b}}=\frac{{z}}{{c}}={k}\:\:\left[{k}\neq\mathrm{0}\right] \\ $$$$\Rightarrow{x}={ak},{y}={bk}\:,{z}={ck} \\ $$$$\frac{{k}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{{k}\left({a}+{b}\right)\left({a}−{b}\right)}\:+\frac{{k}\left({b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)}{{k}\left({b}+{c}\right)\left({b}−{c}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:+\frac{{k}\left({c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}{{k}\left({c}+{a}\right)\left({c}−{a}\right)} \\ $$$$=\frac{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}\:+\frac{\left({b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)}{\left({b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)}\:+\frac{\left({c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}{\left({c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)} \\ $$$$=\mathrm{1}+\mathrm{1}+\mathrm{1}=\mathrm{3} \\ $$

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