Question Number 76436 by Master last updated on 27/Dec/19
Answered by john santu last updated on 27/Dec/19
$$\frac{\mathrm{1}}{{abc}}+\frac{{a}+{c}}{{ac}}=\frac{\mathrm{1}}{{b}} \\ $$$$\frac{{ab}+{bc}+\mathrm{1}}{{abc}}=\frac{\mathrm{1}}{{b}}\rightarrow{ab}+{bc}+\mathrm{1}={ac} \\ $$$${ab}+{bc}={ac}−\mathrm{1}\:\rightarrow{b}=\frac{{ac}−\mathrm{1}}{{a}+{c}} \\ $$$${by}\:{substitusi} \\ $$$$\frac{{a}+{c}}{{ac}\left({ac}−\mathrm{1}\right)}+\frac{{a}+{c}}{{ac}}=\frac{{a}+{c}}{{ac}−\mathrm{1}} \\ $$$$\frac{\left({a}+{c}\right)\left(\mathrm{1}+{ac}−\mathrm{1}\right)}{{ac}\left({ac}−\mathrm{1}\right)}=\frac{{a}+{c}}{{ac}−\mathrm{1}} \\ $$
Commented by john santu last updated on 27/Dec/19
$$\frac{{ac}\left({a}+{c}\right)}{{ac}\left({ac}−\mathrm{1}\right)}=\frac{{ac}\left({a}+{c}\right)}{{ac}\left({ac}−\mathrm{1}\right)}??? \\ $$
Answered by MJS last updated on 27/Dec/19
![11(4/7) solve the equation for c, insert into the expression, then solve (d/da)[exp.]=0 for a, insert into the exp. and solve (d/db)[exp.]=0 for b ⇒ b=±((√5)/3), a=∓((√5)/3), c=((2(√5))/(15))](https://www.tinkutara.com/question/Q76457.png)
$$\mathrm{11}\frac{\mathrm{4}}{\mathrm{7}} \\ $$$$\mathrm{solve}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{for}\:{c},\:\mathrm{insert}\:\mathrm{into}\:\mathrm{the} \\ $$$$\mathrm{expression},\:\mathrm{then}\:\mathrm{solve}\:\frac{{d}}{{da}}\left[\mathrm{exp}.\right]=\mathrm{0}\:\mathrm{for}\:{a}, \\ $$$$\mathrm{insert}\:\mathrm{into}\:\mathrm{the}\:\mathrm{exp}.\:\mathrm{and}\:\mathrm{solve}\:\frac{{d}}{{db}}\left[\mathrm{exp}.\right]=\mathrm{0} \\ $$$$\mathrm{for}\:{b} \\ $$$$\Rightarrow\:{b}=\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{3}},\:{a}=\mp\frac{\sqrt{\mathrm{5}}}{\mathrm{3}},\:{c}=\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{15}} \\ $$
Commented by MJS last updated on 28/Dec/19
![knowing (at least I know from my experience) that a^2 and b^2 are equal in the given expression it′s either a=b or a=−b then it′s even easier (1) a=b (1/(abc))+(1/a)+(1/c)=(1/b) → (1/(b^2 c))+(1/b)+(1/c)=(1/b) ⇒ c is not defined (2) a=−b (1/(abc))+(1/a)+(1/c)=(1/b) → −(1/(b^2 c))−(1/b)+(1/c)=(1/b) ⇒ c=((b^2 −1)/(2b)) (4/(a^2 +1))+(4/(b^2 +1))+(7/(c^2 +1)) → (8/(b^2 +1))+(7/(c^2 +1))= =((4(9b^2 +2))/((b^2 +1)^2 )) (d/db)[((4(9b^2 +2))/((b^2 +1)^2 ))]=0 −((8b(9b^2 −5))/((b^2 +1)^3 ))=0 ⇒ b=0∨b=±((√5)/3) and the rest is obvious](https://www.tinkutara.com/question/Q76519.png)
$$\mathrm{knowing}\:\left(\mathrm{at}\:\mathrm{least}\:\mathrm{I}\:\mathrm{know}\:\mathrm{from}\:\mathrm{my}\:\mathrm{experience}\right) \\ $$$$\mathrm{that}\:{a}^{\mathrm{2}} \:\mathrm{and}\:{b}^{\mathrm{2}} \:\mathrm{are}\:\mathrm{equal}\:\mathrm{in}\:\mathrm{the}\:\mathrm{given}\:\mathrm{expression} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{either}\:{a}={b}\:\mathrm{or}\:{a}=−{b} \\ $$$$\mathrm{then}\:\mathrm{it}'\mathrm{s}\:\mathrm{even}\:\mathrm{easier} \\ $$$$\left(\mathrm{1}\right)\:{a}={b} \\ $$$$\frac{\mathrm{1}}{{abc}}+\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{{b}}\:\rightarrow\:\frac{\mathrm{1}}{{b}^{\mathrm{2}} {c}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{{b}}\:\Rightarrow\:{c}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined} \\ $$$$\left(\mathrm{2}\right)\:{a}=−{b} \\ $$$$\frac{\mathrm{1}}{{abc}}+\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{{b}}\:\rightarrow\:−\frac{\mathrm{1}}{{b}^{\mathrm{2}} {c}}−\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{{b}}\:\Rightarrow\:{c}=\frac{{b}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{b}} \\ $$$$\frac{\mathrm{4}}{{a}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{4}}{{b}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{7}}{{c}^{\mathrm{2}} +\mathrm{1}}\:\rightarrow\:\frac{\mathrm{8}}{{b}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{7}}{{c}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\frac{\mathrm{4}\left(\mathrm{9}{b}^{\mathrm{2}} +\mathrm{2}\right)}{\left({b}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\frac{{d}}{{db}}\left[\frac{\mathrm{4}\left(\mathrm{9}{b}^{\mathrm{2}} +\mathrm{2}\right)}{\left({b}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\right]=\mathrm{0} \\ $$$$−\frac{\mathrm{8}{b}\left(\mathrm{9}{b}^{\mathrm{2}} −\mathrm{5}\right)}{\left({b}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} }=\mathrm{0}\:\Rightarrow\:{b}=\mathrm{0}\vee{b}=\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{3}} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{rest}\:\mathrm{is}\:\mathrm{obvious} \\ $$
Commented by benjo last updated on 28/Dec/19
$$\mathrm{please}\:\mathrm{step}\:\mathrm{by}\:\mathrm{step}\:\mathrm{sir} \\ $$
Commented by MJS last updated on 28/Dec/19
![(1/(abc))+(1/a)+(1/c)=(1/b) ⇒ a≠0∧b≠0∧c≠0∧c=((ab+1)/(a−b)) (4/(a^2 +1))+(4/(b^2 +1))+(7/(c^2 +1))= =((11a^2 −14ab+11b^2 +8)/((a^2 +1)(b^2 +1))) (d/da)[((11a^2 −14ab+11b^2 +8)/((a^2 +1)(b^2 +1)))]=0 ((14ba^2 −2(11b^2 −3)a−14b)/((a^2 +1)^2 (b^2 +1)))=0 ⇒ a=((11b^2 −3±(√(121b^4 +130b^2 +9)))/(14b)) ((11a^2 −14ab+11b^2 +8)/((a^2 +1)(b^2 +1)))= =((11b^2 +19±(√(121b^4 +130b^2 +9)))/(2(b^2 +1))) (d/db)[((11b^2 +19±(√(121b^4 +130b^2 +9)))/(2(b^2 +1)))]=0 −((8b(±7(b^2 +1)+(√(121b^4 +130b^2 +9)))/((b^2 +1)^2 (√(121b^4 +130b^2 +9))))=0 ⇒ b=0∨b=±((√5)/3) but b≠0 ⇒ a=∓((√5)/3), c=((2(√5))/(15)) I hope you are able to solve the equations, I don′t want to type all the steps](https://www.tinkutara.com/question/Q76517.png)
$$\frac{\mathrm{1}}{{abc}}+\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{{b}}\:\Rightarrow\:{a}\neq\mathrm{0}\wedge{b}\neq\mathrm{0}\wedge{c}\neq\mathrm{0}\wedge{c}=\frac{{ab}+\mathrm{1}}{{a}−{b}} \\ $$$$\frac{\mathrm{4}}{{a}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{4}}{{b}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{7}}{{c}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\frac{\mathrm{11}{a}^{\mathrm{2}} −\mathrm{14}{ab}+\mathrm{11}{b}^{\mathrm{2}} +\mathrm{8}}{\left({a}^{\mathrm{2}} +\mathrm{1}\right)\left({b}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\frac{{d}}{{da}}\left[\frac{\mathrm{11}{a}^{\mathrm{2}} −\mathrm{14}{ab}+\mathrm{11}{b}^{\mathrm{2}} +\mathrm{8}}{\left({a}^{\mathrm{2}} +\mathrm{1}\right)\left({b}^{\mathrm{2}} +\mathrm{1}\right)}\right]=\mathrm{0} \\ $$$$\frac{\mathrm{14}{ba}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{11}{b}^{\mathrm{2}} −\mathrm{3}\right){a}−\mathrm{14}{b}}{\left({a}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \left({b}^{\mathrm{2}} +\mathrm{1}\right)}=\mathrm{0} \\ $$$$\Rightarrow\:{a}=\frac{\mathrm{11}{b}^{\mathrm{2}} −\mathrm{3}\pm\sqrt{\mathrm{121}{b}^{\mathrm{4}} +\mathrm{130}{b}^{\mathrm{2}} +\mathrm{9}}}{\mathrm{14}{b}} \\ $$$$\frac{\mathrm{11}{a}^{\mathrm{2}} −\mathrm{14}{ab}+\mathrm{11}{b}^{\mathrm{2}} +\mathrm{8}}{\left({a}^{\mathrm{2}} +\mathrm{1}\right)\left({b}^{\mathrm{2}} +\mathrm{1}\right)}= \\ $$$$=\frac{\mathrm{11}{b}^{\mathrm{2}} +\mathrm{19}\pm\sqrt{\mathrm{121}{b}^{\mathrm{4}} +\mathrm{130}{b}^{\mathrm{2}} +\mathrm{9}}}{\mathrm{2}\left({b}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\frac{{d}}{{db}}\left[\frac{\mathrm{11}{b}^{\mathrm{2}} +\mathrm{19}\pm\sqrt{\mathrm{121}{b}^{\mathrm{4}} +\mathrm{130}{b}^{\mathrm{2}} +\mathrm{9}}}{\mathrm{2}\left({b}^{\mathrm{2}} +\mathrm{1}\right)}\right]=\mathrm{0} \\ $$$$−\frac{\mathrm{8}{b}\left(\pm\mathrm{7}\left({b}^{\mathrm{2}} +\mathrm{1}\right)+\sqrt{\mathrm{121}{b}^{\mathrm{4}} +\mathrm{130}{b}^{\mathrm{2}} +\mathrm{9}}\right.}{\left({b}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \sqrt{\mathrm{121}{b}^{\mathrm{4}} +\mathrm{130}{b}^{\mathrm{2}} +\mathrm{9}}}=\mathrm{0} \\ $$$$\Rightarrow\:{b}=\mathrm{0}\vee{b}=\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{3}}\:\mathrm{but}\:{b}\neq\mathrm{0} \\ $$$$\Rightarrow\:{a}=\mp\frac{\sqrt{\mathrm{5}}}{\mathrm{3}},\:{c}=\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{15}} \\ $$$$\mathrm{I}\:\mathrm{hope}\:\mathrm{you}\:\mathrm{are}\:\mathrm{able}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{equations}, \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{want}\:\mathrm{to}\:\mathrm{type}\:\mathrm{all}\:\mathrm{the}\:\mathrm{steps} \\ $$
Commented by john santu last updated on 28/Dec/19
$${oo}\:{yes}.\:{thanks}\:{sir} \\ $$
Commented by Master last updated on 28/Dec/19
$$\mathrm{thanks} \\ $$