Question Number 185876 by MATHEMATICSAM last updated on 29/Jan/23
$${x}\:=\:\frac{\mathrm{2}{ab}}{{a}+{b}}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\: \\ $$$$\frac{{x}\:+\:{a}}{{x}\:−\:{a}}\:+\:\frac{{x}\:+\:{b}}{{x}\:−\:{b}}\:=\:\mathrm{2} \\ $$
Answered by Rasheed.Sindhi last updated on 29/Jan/23
$${x}\:=\:\frac{\mathrm{2}{ab}}{{a}+{b}}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\: \\ $$$$\frac{{x}\:+\:{a}}{{x}\:−\:{a}}\:+\:\frac{{x}\:+\:{b}}{{x}\:−\:{b}}\:=\:\mathrm{2} \\ $$$$\: \\ $$$$\mathrm{LHS}: \\ $$$$=\frac{{x}\:+\:{a}}{{x}\:−\:{a}}−\mathrm{1}\:+\:\frac{{x}\:+\:{b}}{{x}\:−\:{b}}−\mathrm{1}+\mathrm{2} \\ $$$$=\frac{\mathrm{2}{a}}{{x}−{a}}+\frac{\mathrm{2}{b}}{{x}−{b}}+\mathrm{2} \\ $$$$=\frac{\mathrm{2}{a}}{\frac{\mathrm{2}{ab}}{{a}+{b}}−{a}}+\frac{\mathrm{2}{b}}{\frac{\mathrm{2}{ab}}{{a}+{b}}−{b}}+\mathrm{2} \\ $$$$=\frac{\mathrm{2}{a}}{\frac{\mathrm{2}{ab}−{a}^{\mathrm{2}} −{ab}}{{a}+{b}}}+\frac{\mathrm{2}{b}}{\frac{\mathrm{2}{ab}−{ab}−{b}^{\mathrm{2}} }{{a}+{b}}}+\mathrm{2} \\ $$$$=\frac{\mathrm{2}}{\frac{{b}−{a}}{{a}+{b}}}+\frac{\mathrm{2}}{\frac{{a}−{b}}{{a}+{b}}}+\mathrm{2} \\ $$$$=−\cancel{\frac{\mathrm{2}\left({a}+{b}\right)}{{a}−{b}}}+\cancel{\frac{\mathrm{2}\left({a}+{b}\right)}{{a}−{b}}}+\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}=\mathrm{RHS} \\ $$$$\:\:\:\:\:\:\:\mathrm{Proved} \\ $$
Answered by ajfour last updated on 29/Jan/23
$${Assuming}\:{true}\:{that}\:{which}\:{to} \\ $$$${prove}; \\ $$$$\left({x}+{a}\right)\left({x}−{b}\right)+\left({x}−{a}\right)\left({x}+{b}\right) \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{2}\left({x}−{a}\right)\left({x}−{b}\right) \\ $$$$\Rightarrow\:−\mathrm{2}{ab}=−\mathrm{2}\left({a}+{b}\right){x}+\mathrm{2}{ab} \\ $$$$\Rightarrow\:\:\:{x}=\frac{\mathrm{2}{ab}}{{a}+{b}}\:\:\left({so}\:{must}\:{be}\right) \\ $$
Answered by HeferH last updated on 29/Jan/23
$$\:{i}.\:{x}\left({a}\:+{b}\right)\:=\:\mathrm{2}{ab} \\ $$$$\:{ii}.\:\frac{{x}\:+\:{a}}{{x}\:−{a}}\:+\:\frac{{x}\:+\:{b}}{{x}\:−\:{b}}\:=\:\mathrm{2}\:+\:\mathrm{2}\left(\frac{{a}}{{x}\:−\:{a}}\:+\:\frac{{b}}{{x}\:−\:{b}}\right)\: \\ $$$$\:=\:\mathrm{2}\:+\:\mathrm{2}\left[\frac{{a}\left({x}−{b}\right)\:+\:{b}\left({x}−{a}\right)}{\left({x}−{a}\right)\left({x}−{b}\right)}\right] \\ $$$$\:=\:\mathrm{2}\:+\:\mathrm{2}\left[\frac{{x}\left({a}+{b}\right)−\mathrm{2}{ab}}{\left({x}−{a}\right)\left({x}−{b}\right)}\right]\: \\ $$$$\:=\:\mathrm{2}\:+\:\mathrm{2}\left[\frac{\mathrm{0}}{\left({x}−{a}\right)\left({x}−{b}\right)}\right]\:=\:\mathrm{2}\:+\:\mathrm{2}\left(\mathrm{0}\right)\:=\:\mathrm{2}\:\checkmark \\ $$