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Question Number 200242 by cherokeesay last updated on 16/Nov/23

Answered by cortano12 last updated on 17/Nov/23

(((ab+c)x)/(b−(c/a) )) −(((ab−c)x)/(b+(c/a))) = ((ab−c)/(b+(c/a))) −((ab+c)/(b−(c/a))) ((ax(ab+c))/(ab−c)) −((ax(ab−c))/(ab+c)) = ((a(ab−c))/(ab+c)) −((a(ab+c))/(ab−c)) ((x(ab+c))/(ab−c)) −((x(ab−c))/(ab+c)) = ((ab−c)/(ab+c))−((ab+c)/(ab−c)) (((ab+c)(x+1))/(ab−c)) = (((ab−c)(x+1))/(ab+c)) (ab+c)^2 (x+1)= (ab−c)^2 (x+1) (x+1){(ab+c)^2 −(ab−c)^2 } = 0 (x+1)(2ab)(2c)=0 x = −1

(ab+c)xb−ca−(ab−c)xb+ca=ab−cb+ca−ab+cb−caax(ab+c)ab−c−ax(ab−c)ab+c=a(ab−c)ab+c−a(ab+c)ab−cx(ab+c)ab−c−x(ab−c)ab+c=ab−cab+c−ab+cab−c(ab+c)(x+1)ab−c=(ab−c)(x+1)ab+c(ab+c)2(x+1)=(ab−c)2(x+1)(x+1){(ab+c)2−(ab−c)2}=0(x+1)(2ab)(2c)=0x=−1

Commented by cherokeesay last updated on 16/Nov/23

thanks !

thanks!

Answered by som(math1967) last updated on 16/Nov/23

((ax(ab+c))/(ab−c)) +((a(ab+c))/(ab−c))=((ax(ab−c))/(ab+c))+((a(ab−c))/(ab+c)) ⇒ax{(((ab+c)^2 −(ab−c)^2 )/(a^2 b^2 −c^2 ))}=a{(((ab−c)^2 −(ab+c)^2 )/(a^2 b^2 −c^2 ))} x=((−a)/a)=−1

ax(ab+c)ab−c+a(ab+c)ab−c=ax(ab−c)ab+c+a(ab−c)ab+c⇒ax{(ab+c)2−(ab−c)2a2b2−c2}=a{(ab−c)2−(ab+c)2a2b2−c2}x=−aa=−1

Commented by cherokeesay last updated on 16/Nov/23

thank you !

thankyou!

Answered by Rasheed.Sindhi last updated on 16/Nov/23

(((ab+c)x)/(b−(c/a)))−((ab−c)/(b+(c/a)))=(((ab−c)x)/(b+(c/a)))−((ab+c)/(b−(c/a))) (((ab+c)/(b−(c/a)))−((ab−c)/(b+(c/a))))x=((ab−c)/(b+(c/a)))−((ab+c)/(b−(c/a))) a(((b+(c/a))/(b−(c/a)))−((b−(c/a))/(b+(c/a))))x=−a(((b+(c/a))/(b−(c/a)))−((b−(c/a))/(b+(c/a)))) [a≠0] x=−1

(ab+c)xb−ca−ab−cb+ca=(ab−c)xb+ca−ab+cb−ca(ab+cb−ca−ab−cb+ca)x=ab−cb+ca−ab+cb−caa(b+cab−ca−b−cab+ca)x=−a(b+cab−ca−b−cab+ca)[a≠0]x=−1

Commented by cherokeesay last updated on 16/Nov/23

thank you !

thankyou!

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