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Question Number 185876 by MATHEMATICSAM last updated on 29/Jan/23

x = ((2ab)/(a+b)) then prove that ((x + a)/(x − a)) + ((x + b)/(x − b)) = 2

$${x}\:=\:\frac{\mathrm{2}{ab}}{{a}+{b}}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\: \\ $$$$\frac{{x}\:+\:{a}}{{x}\:−\:{a}}\:+\:\frac{{x}\:+\:{b}}{{x}\:−\:{b}}\:=\:\mathrm{2} \\ $$

Answered by Rasheed.Sindhi last updated on 29/Jan/23

x = ((2ab)/(a+b)) then prove that ((x + a)/(x − a)) + ((x + b)/(x − b)) = 2 LHS: =((x + a)/(x − a))−1 + ((x + b)/(x − b))−1+2 =((2a)/(x−a))+((2b)/(x−b))+2 =((2a)/(((2ab)/(a+b))−a))+((2b)/(((2ab)/(a+b))−b))+2 =((2a)/((2ab−a^2 −ab)/(a+b)))+((2b)/((2ab−ab−b^2 )/(a+b)))+2 =(2/((b−a)/(a+b)))+(2/((a−b)/(a+b)))+2 =−((2(a+b))/(a−b))+((2(a+b))/(a−b))+2 =2=RHS Proved

$${x}\:=\:\frac{\mathrm{2}{ab}}{{a}+{b}}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\: \\ $$$$\frac{{x}\:+\:{a}}{{x}\:−\:{a}}\:+\:\frac{{x}\:+\:{b}}{{x}\:−\:{b}}\:=\:\mathrm{2} \\ $$$$\: \\ $$$$\mathrm{LHS}: \\ $$$$=\frac{{x}\:+\:{a}}{{x}\:−\:{a}}−\mathrm{1}\:+\:\frac{{x}\:+\:{b}}{{x}\:−\:{b}}−\mathrm{1}+\mathrm{2} \\ $$$$=\frac{\mathrm{2}{a}}{{x}−{a}}+\frac{\mathrm{2}{b}}{{x}−{b}}+\mathrm{2} \\ $$$$=\frac{\mathrm{2}{a}}{\frac{\mathrm{2}{ab}}{{a}+{b}}−{a}}+\frac{\mathrm{2}{b}}{\frac{\mathrm{2}{ab}}{{a}+{b}}−{b}}+\mathrm{2} \\ $$$$=\frac{\mathrm{2}{a}}{\frac{\mathrm{2}{ab}−{a}^{\mathrm{2}} −{ab}}{{a}+{b}}}+\frac{\mathrm{2}{b}}{\frac{\mathrm{2}{ab}−{ab}−{b}^{\mathrm{2}} }{{a}+{b}}}+\mathrm{2} \\ $$$$=\frac{\mathrm{2}}{\frac{{b}−{a}}{{a}+{b}}}+\frac{\mathrm{2}}{\frac{{a}−{b}}{{a}+{b}}}+\mathrm{2} \\ $$$$=−\cancel{\frac{\mathrm{2}\left({a}+{b}\right)}{{a}−{b}}}+\cancel{\frac{\mathrm{2}\left({a}+{b}\right)}{{a}−{b}}}+\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}=\mathrm{RHS} \\ $$$$\:\:\:\:\:\:\:\mathrm{Proved} \\ $$

Answered by ajfour last updated on 29/Jan/23

Assuming true that which to prove; (x+a)(x−b)+(x−a)(x+b) =2(x−a)(x−b) ⇒ −2ab=−2(a+b)x+2ab ⇒ x=((2ab)/(a+b)) (so must be)

$${Assuming}\:{true}\:{that}\:{which}\:{to} \\ $$$${prove}; \\ $$$$\left({x}+{a}\right)\left({x}−{b}\right)+\left({x}−{a}\right)\left({x}+{b}\right) \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{2}\left({x}−{a}\right)\left({x}−{b}\right) \\ $$$$\Rightarrow\:−\mathrm{2}{ab}=−\mathrm{2}\left({a}+{b}\right){x}+\mathrm{2}{ab} \\ $$$$\Rightarrow\:\:\:{x}=\frac{\mathrm{2}{ab}}{{a}+{b}}\:\:\left({so}\:{must}\:{be}\right) \\ $$

Answered by HeferH last updated on 29/Jan/23

i. x(a +b) = 2ab ii. ((x + a)/(x −a)) + ((x + b)/(x − b)) = 2 + 2((a/(x − a)) + (b/(x − b))) = 2 + 2[((a(x−b) + b(x−a))/((x−a)(x−b)))] = 2 + 2[((x(a+b)−2ab)/((x−a)(x−b)))] = 2 + 2[(0/((x−a)(x−b)))] = 2 + 2(0) = 2 ✓

$$\:{i}.\:{x}\left({a}\:+{b}\right)\:=\:\mathrm{2}{ab} \\ $$$$\:{ii}.\:\frac{{x}\:+\:{a}}{{x}\:−{a}}\:+\:\frac{{x}\:+\:{b}}{{x}\:−\:{b}}\:=\:\mathrm{2}\:+\:\mathrm{2}\left(\frac{{a}}{{x}\:−\:{a}}\:+\:\frac{{b}}{{x}\:−\:{b}}\right)\: \\ $$$$\:=\:\mathrm{2}\:+\:\mathrm{2}\left[\frac{{a}\left({x}−{b}\right)\:+\:{b}\left({x}−{a}\right)}{\left({x}−{a}\right)\left({x}−{b}\right)}\right] \\ $$$$\:=\:\mathrm{2}\:+\:\mathrm{2}\left[\frac{{x}\left({a}+{b}\right)−\mathrm{2}{ab}}{\left({x}−{a}\right)\left({x}−{b}\right)}\right]\: \\ $$$$\:=\:\mathrm{2}\:+\:\mathrm{2}\left[\frac{\mathrm{0}}{\left({x}−{a}\right)\left({x}−{b}\right)}\right]\:=\:\mathrm{2}\:+\:\mathrm{2}\left(\mathrm{0}\right)\:=\:\mathrm{2}\:\checkmark \\ $$

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