Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 148428 by 0731619 last updated on 27/Jul/21

Answered by Mathspace last updated on 27/Jul/21

(((a+b)^2 )/(ab))−3 =(((a+b)^2 −3ab)/(ab))  =((a^2 +2ab+b^2 −3ab)/(ab))  =((a^2 −ab+b^2 )/(ab))=((a^2 −2a(b/2)+(b^2 /4)+b^2 −(b^2 /4))/(ab))  =(((a−(b/(2 )))^2 +((3b^2 )/4))/(ab))>0

$$\frac{\left({a}+{b}\right)^{\mathrm{2}} }{{ab}}−\mathrm{3}\:=\frac{\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{3}{ab}}{{ab}} \\ $$$$=\frac{{a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} −\mathrm{3}{ab}}{{ab}} \\ $$$$=\frac{{a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} }{{ab}}=\frac{{a}^{\mathrm{2}} −\mathrm{2}{a}\frac{{b}}{\mathrm{2}}+\frac{{b}^{\mathrm{2}} }{\mathrm{4}}+{b}^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{\mathrm{4}}}{{ab}} \\ $$$$=\frac{\left({a}−\frac{{b}}{\mathrm{2}\:}\right)^{\mathrm{2}} +\frac{\mathrm{3}{b}^{\mathrm{2}} }{\mathrm{4}}}{{ab}}>\mathrm{0} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com