Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 5022 by love math last updated on 03/Apr/16

((n^(1.2) −3(√3))/(n^(0.8) −(√(3n^(0.8) ))+3))−(√3)((√(3n^(0.8) ))−1) Simplify

$$\frac{{n}^{\mathrm{1}.\mathrm{2}} −\mathrm{3}\sqrt{\mathrm{3}}}{{n}^{\mathrm{0}.\mathrm{8}} −\sqrt{\mathrm{3}{n}^{\mathrm{0}.\mathrm{8}} }+\mathrm{3}}−\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}{n}^{\mathrm{0}.\mathrm{8}} }−\mathrm{1}\right) \\ $$$$ \\ $$$${Simplify} \\ $$

Commented by prakash jain last updated on 04/Apr/16

Substitute to ease process of simplification n^(.4) =a (√3)=b ((a^3 −b^3 )/(a^2 −ab+b^2 ))−b(ab−1) It looks the question is probably mistyped the first probably was ((n^(1.2) −3(√3))/(n^(0.8) +(√(3n^(0.8) ))+3)) In that case first part simplifies to ((a^3 −b^3 )/(a^2 +ab+b^2 ))−b(ab−1) =(((a−b)(a^2 +ab+b^2 ))/((a^2 +ab+b^2 )))−b(ab−1) =a−b−ab^2 +b =a−ab^2 =a(1−b^2 ) =−2n^(0.4)

$$\mathrm{Substitute}\:\mathrm{to}\:\mathrm{ease}\:\mathrm{process}\:\mathrm{of}\:\mathrm{simplification} \\ $$$${n}^{.\mathrm{4}} ={a} \\ $$$$\sqrt{\mathrm{3}}={b} \\ $$$$\frac{{a}^{\mathrm{3}} −{b}^{\mathrm{3}} }{{a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} }−{b}\left({ab}−\mathrm{1}\right) \\ $$$$\mathrm{It}\:\mathrm{looks}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{probably}\:\mathrm{mistyped} \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{probably}\:\mathrm{was}\:\frac{{n}^{\mathrm{1}.\mathrm{2}} −\mathrm{3}\sqrt{\mathrm{3}}}{{n}^{\mathrm{0}.\mathrm{8}} +\sqrt{\mathrm{3}{n}^{\mathrm{0}.\mathrm{8}} }+\mathrm{3}} \\ $$$$\mathrm{In}\:\mathrm{that}\:\mathrm{case}\:\mathrm{first}\:\mathrm{part}\:\mathrm{simplifies}\:\mathrm{to} \\ $$$$\frac{{a}^{\mathrm{3}} −{b}^{\mathrm{3}} }{{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} }−{b}\left({ab}−\mathrm{1}\right) \\ $$$$=\frac{\left({a}−{b}\right)\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right)}{\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right)}−{b}\left({ab}−\mathrm{1}\right) \\ $$$$={a}−{b}−{ab}^{\mathrm{2}} +{b} \\ $$$$={a}−{ab}^{\mathrm{2}} \\ $$$$={a}\left(\mathrm{1}−{b}^{\mathrm{2}} \right) \\ $$$$=−\mathrm{2}{n}^{\mathrm{0}.\mathrm{4}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com