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Question Number 187076 by Humble last updated on 13/Feb/23
(a/b)+(b/a)=5 ; (a^2 /b)+(b^2 /a) =12 (1/a)+(1/b)=?
$$\frac{{a}}{{b}}+\frac{{b}}{{a}}=\mathrm{5}\:\:;\:\frac{{a}^{\mathrm{2}} }{{b}}+\frac{{b}^{\mathrm{2}} }{{a}}\:=\mathrm{12} \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=? \\ $$
Answered by Rasheed.Sindhi last updated on 13/Feb/23
(a/b)+(b/a)=5 ; (a^2 /b)+(b^2 /a) =12 ;(1/a)+(1/b)=? a^2 +b^2 =5ab ; a^3 +b^3 =12ab (a+b)^2 =7ab ; (a+b)^3 =12ab+3ab(a+b) a+b=((ab(12+3(a+b)))/(7ab)) a+b=((12)/7)+(3/7)(a+b) 7(a+b)−3(a+b)=12 a+b=12/4=3 ab=(((a+b)^2 )/7)=(3^2 /7)=(9/7) ((a+b)/(ab))=(3/(9/7))=3×(7/9)=(7/3) (1/a)+(1/b)=(7/3)
$$\frac{{a}}{{b}}+\frac{{b}}{{a}}=\mathrm{5}\:\:;\:\frac{{a}^{\mathrm{2}} }{{b}}+\frac{{b}^{\mathrm{2}} }{{a}}\:=\mathrm{12}\:\:;\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=? \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{5}{ab}\:;\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\mathrm{12}{ab} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} =\mathrm{7}{ab}\:;\:\left({a}+{b}\right)^{\mathrm{3}} =\mathrm{12}{ab}+\mathrm{3}{ab}\left({a}+{b}\right) \\ $$$${a}+{b}=\frac{{ab}\left(\mathrm{12}+\mathrm{3}\left({a}+{b}\right)\right)}{\mathrm{7}{ab}} \\ $$$${a}+{b}=\frac{\mathrm{12}}{\mathrm{7}}+\frac{\mathrm{3}}{\mathrm{7}}\left({a}+{b}\right) \\ $$$$\mathrm{7}\left({a}+{b}\right)−\mathrm{3}\left({a}+{b}\right)=\mathrm{12} \\ $$$${a}+{b}=\mathrm{12}/\mathrm{4}=\mathrm{3} \\ $$$${ab}=\frac{\left({a}+{b}\right)^{\mathrm{2}} }{\mathrm{7}}=\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{7}}=\frac{\mathrm{9}}{\mathrm{7}} \\ $$$$\frac{{a}+{b}}{{ab}}=\frac{\mathrm{3}}{\mathrm{9}/\mathrm{7}}=\mathrm{3}×\frac{\mathrm{7}}{\mathrm{9}}=\frac{\mathrm{7}}{\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{\mathrm{7}}{\mathrm{3}} \\ $$
Commented by BaliramKumar last updated on 13/Feb/23
step3 devid will be 7ab
$${step}\mathrm{3} \\ $$$${devid}\:{will}\:{be}\:\mathrm{7}{ab} \\ $$
Commented by Rasheed.Sindhi last updated on 13/Feb/23
ThanX Baliram.I′ve corrected.
$$\mathcal{T}{han}\mathcal{X}\:{Baliram}.{I}'{ve}\:{corrected}. \\ $$
Commented by Humble last updated on 13/Feb/23
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Answered by HeferH last updated on 13/Feb/23
(b/a) + (b/a) = 5 ⇒ a^2 + b^2 = 5ab (a^2 /b) + (b^2 /a) = 12 ⇒ a^3 + b^3 =12ab (a + b)(a^2 + b^2 −ab) = 12ab (a + b)(5ab −ab) = 12ab a + b = 3 (a + b)^2 = 3^2 a^2 + b^(2 ) + 2ab = 9 5ab + 2ab = 9 ab = (9/7) ⇒ (1/a) + (1/b) = ((a + b)/(ab)) = (3/(9/7)) = 3∙(7/(9 )) = (7/3)
$$\:\frac{{b}}{{a}}\:+\:\frac{{b}}{{a}}\:=\:\mathrm{5}\:\Rightarrow \\ $$$$\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:=\:\mathrm{5}{ab} \\ $$$$\:\frac{{a}^{\mathrm{2}} }{{b}}\:+\:\frac{{b}^{\mathrm{2}} }{{a}}\:=\:\mathrm{12}\:\Rightarrow \\ $$$$\:{a}^{\mathrm{3}} \:+\:{b}^{\mathrm{3}} \:=\mathrm{12}{ab} \\ $$$$\:\left({a}\:+\:{b}\right)\left({a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:−{ab}\right)\:=\:\mathrm{12}{ab} \\ $$$$\:\left({a}\:+\:{b}\right)\left(\mathrm{5}{ab}\:−{ab}\right)\:=\:\mathrm{12}{ab} \\ $$$$\:{a}\:+\:{b}\:=\:\mathrm{3} \\ $$$$\:\left({a}\:+\:{b}\right)^{\mathrm{2}} \:=\:\mathrm{3}^{\mathrm{2}} \\ $$$$\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}\:} \:+\:\mathrm{2}{ab}\:=\:\mathrm{9} \\ $$$$\:\mathrm{5}{ab}\:+\:\mathrm{2}{ab}\:=\:\mathrm{9}\: \\ $$$$\:{ab}\:=\:\frac{\mathrm{9}}{\mathrm{7}}\:\Rightarrow \\ $$$$\:\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}}\:=\:\frac{{a}\:+\:{b}}{{ab}}\:=\:\frac{\mathrm{3}}{\frac{\mathrm{9}}{\mathrm{7}}}\:=\:\mathrm{3}\centerdot\frac{\mathrm{7}}{\mathrm{9}\:}\:=\:\frac{\mathrm{7}}{\mathrm{3}} \\ $$
Commented by Humble last updated on 13/Feb/23
thank you, sir
$${thank}\:{you},\:{sir} \\ $$
Answered by Rasheed.Sindhi last updated on 13/Feb/23
(a/b)+(b/a)=5_((i)) ; (a^2 /b)+(b^2 /a) =12_((ii)) ; (1/a)+(1/b)=? (i)⇒a^2 +b^2 =5ab (ii)⇒a^3 +b^3 =12ab (i)⇒a^2 −ab+b^2 =4ab (a+b)(a^2 −ab+b^2 =4ab(a+b) a^3 +b^3 =4ab(a+b) 12ab=4ab(a+b) a+b=3....(iii) a^2 +b^2 =5ab a^2 +2ab+b^2 =7ab ab=(((a+b)^2 )/7)=(3^2 /7)=(9/7)....(iv) (iii)/(iv): ((a+b)/(ab))=3/((9/7))=3×(7/9)=(7/3) (1/a)+(1/b)=(7/3)
$$\underset{\left({i}\right)} {\underbrace{\frac{{a}}{{b}}+\frac{{b}}{{a}}=\mathrm{5}}}\:\:;\:\underset{\left({ii}\right)} {\underbrace{\frac{{a}^{\mathrm{2}} }{{b}}+\frac{{b}^{\mathrm{2}} }{{a}}\:=\mathrm{12}}}\:;\:\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=? \\ $$$$\left({i}\right)\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{5}{ab}\: \\ $$$$\left({ii}\right)\Rightarrow{a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\mathrm{12}{ab} \\ $$$$\left({i}\right)\Rightarrow{a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} =\mathrm{4}{ab} \\ $$$$\left({a}+{b}\right)\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} =\mathrm{4}{ab}\left({a}+{b}\right)\right. \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\mathrm{4}{ab}\left({a}+{b}\right) \\ $$$$\mathrm{12}{ab}=\mathrm{4}{ab}\left({a}+{b}\right) \\ $$$${a}+{b}=\mathrm{3}….\left({iii}\right) \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{5}{ab} \\ $$$${a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} =\mathrm{7}{ab} \\ $$$${ab}=\frac{\left({a}+{b}\right)^{\mathrm{2}} }{\mathrm{7}}=\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{7}}=\frac{\mathrm{9}}{\mathrm{7}}….\left({iv}\right) \\ $$$$\left({iii}\right)/\left({iv}\right):\:\frac{{a}+{b}}{{ab}}=\mathrm{3}/\left(\frac{\mathrm{9}}{\mathrm{7}}\right)=\mathrm{3}×\frac{\mathrm{7}}{\mathrm{9}}=\frac{\mathrm{7}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{\mathrm{7}}{\mathrm{3}} \\ $$
Commented by Humble last updated on 14/Feb/23
thank you, sir
$${thank}\:{you},\:{sir} \\ $$

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