Question Number 186163 by Rupesh123 last updated on 01/Feb/23
Answered by Frix last updated on 01/Feb/23
$${x}=\pm\mathrm{4} \\ $$
Answered by aba last updated on 01/Feb/23
$$\mathrm{x}=\pm\mathrm{4} \\ $$
Answered by manolito last updated on 02/Feb/23
![5+x=a^(2 ) 5−x=b^2 a^2 +b^2 =10 a^2 −b^2 =2x x=((a^2 −b^2 )/2) ((3+((a^2 −b^2 )/2))/a)+((3−((a^2 −b^2 )/2))/b)=(4/3) usar a^2 +b^2 =10 en el desarrollo 3(a+b)[ab−2]=4ab a=3 5+x=9 x=4 b=±1 b=3 5−x=9 x=−4 a=±1 a=a+b b=0 false continuara......... (/)](https://www.tinkutara.com/question/Q186266.png)
$$\mathrm{5}+{x}={a}^{\mathrm{2}\:\:\:\:\:\:\:} \mathrm{5}−{x}={b}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{10}\:\:\:\:\:\:\:{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{2}{x}\:\:\:{x}=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\frac{\mathrm{3}+\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}}}{{a}}+\frac{\mathrm{3}−\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}}}{{b}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${usar}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{10}\:{en}\:{el}\:{desarrollo} \\ $$$$\mathrm{3}\left({a}+{b}\right)\left[{ab}−\mathrm{2}\right]=\mathrm{4}{ab} \\ $$$${a}=\mathrm{3}\:\:\:\:\:\mathrm{5}+{x}=\mathrm{9}\:\:\:\:{x}=\mathrm{4}\:\:\:\:\:{b}=\pm\mathrm{1} \\ $$$${b}=\mathrm{3}\:\:\:\:\:\:\mathrm{5}−{x}=\mathrm{9}\:\:\:{x}=−\mathrm{4}\:\:\:{a}=\pm\mathrm{1} \\ $$$${a}={a}+{b}\:\:\:\:\:{b}=\mathrm{0}\:\:\:\:\:\:\:{false} \\ $$$${continuara}……… \\ $$$$ \\ $$$$\frac{}{} \\ $$