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Question Number 37037 by jayanta11 last updated on 08/Jun/18

If ((a+b−c)/(a+b)) = ((b+c−a)/(b+c)) =((c+a−b)/(c+a)) and a+b+c ≠ 0 then which of the following is true 1) a=−b=c 2)−a=−b=c 3) a=b=c 4) a=b≠c

Ifa+b−ca+b=b+c−ab+c=c+a−bc+aanda+b+c≠0thenwhichofthefollowingistrue1)a=−b=c2)−a=−b=c3)a=b=c4)a=b≠c

Answered by math1967 last updated on 08/Jun/18

3)a=b=c ((a+b)/(a+b))−(c/(a+b))=((b+c)/(b+c))−(a/(b+c))=((c+a)/(c+a))−(b/(c+a)) (c/(a+b))=(a/(b+c))=(b/(c+a)) ((a+b)/c)+1=((b+c)/a)+1=((c+a)/b)+1 ((a+b+c)/c)=((a+b+c)/a)=((a+b+c)/b) (1/c)=(1/a)=(1/b) [(a+b+c)≠0] ∴c=a=b⇒a=b=c

3)a=b=ca+ba+b−ca+b=b+cb+c−ab+c=c+ac+a−bc+aca+b=ab+c=bc+aa+bc+1=b+ca+1=c+ab+1a+b+cc=a+b+ca=a+b+cb1c=1a=1b[(a+b+c)≠0]∴c=a=b⇒a=b=c

Commented by Rasheed.Sindhi last updated on 08/Jun/18

Nice!

Nice!

Answered by Rasheed.Sindhi last updated on 08/Jun/18

((a+b−c)/(a+b)) = ((b+c−a)/(b+c)) =((c+a−b)/(c+a)) =(((a+b−c)+(b+c−a)+(c+a−b))/((a+b)+(b+c)+(c+a))) =((a+b+c)/(2(a+b+c)))=(1/2) 2(a+b−c)=a+b 2(b+c−a)=b+c 2(c+a−b)=c+a a+b−2c=0⇒b=2c−a^∗ b+c−2a=0⇒2c−a+c−2a=0⇒3c=3a^∗ c+a−2b=0⇒c+a−4c+2a=0⇒3a=3c^∗ ^∗ a=b=c

a+b−ca+b=b+c−ab+c=c+a−bc+a=(a+b−c)+(b+c−a)+(c+a−b)(a+b)+(b+c)+(c+a)=a+b+c2(a+b+c)=122(a+b−c)=a+b2(b+c−a)=b+c2(c+a−b)=c+aa+b−2c=0⇒b=2c−a∗b+c−2a=0⇒2c−a+c−2a=0⇒3c=3a∗c+a−2b=0⇒c+a−4c+2a=0⇒3a=3c∗∗a=b=c

Answered by ajfour last updated on 08/Jun/18

((a+b−c)/(a+b))=((b+c−a)/(b+c))=((c+a−b)/(c+a)) ⇒ (c/(a+b))=(a/(b+c))=(b/(c+a)) ⇒ ((a+b+c)/c)=((a+b+c)/a)=((a+b+c)/b) ⇒ a=b=c .

a+b−ca+b=b+c−ab+c=c+a−bc+a⇒ca+b=ab+c=bc+a⇒a+b+cc=a+b+ca=a+b+cb⇒a=b=c.

Answered by tanmay.chaudhury50@gmail.com last updated on 08/Jun/18

((a+b−c)/(a+b))=((b+c−a)/(b+c))=((c+a−b)/(c+a)) 1−(c/(a+b))=1−(a/(b+c))=1−(b/(c+a)) (c/(a+b))=(a/(b+c))=(b/(c+a)) bc+c^2 =a^2 +ab bc−ab+c^2 −a^2 =0 b(c−a)+(c+a)(c−a)=0 (c−a)(b+c+a)=0 since a+b+c not equales to zero so c=a similarly a=b so a=b=c value of each ratio i ((a+b−c)/(a+b)) =((a+a−a)/(a+a)) =(a/(2a))=(1/2) value of each ratio when a+b+c=0 ((a+b−c)/(a+b)) ((−c−c)/(−c)) ((−2c)/(−c)) =2

a+b−ca+b=b+c−ab+c=c+a−bc+a1−ca+b=1−ab+c=1−bc+aca+b=ab+c=bc+abc+c2=a2+abbc−ab+c2−a2=0b(c−a)+(c+a)(c−a)=0(c−a)(b+c+a)=0sincea+b+cnotequalestozerosoc=asimilarlya=bsoa=b=cvalueofeachratioia+b−ca+b=a+a−aa+a=a2a=12valueofeachratiowhena+b+c=0a+b−ca+b−c−c−c−2c−c=2

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