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Question-97479




Question Number 97479 by Hassen_Timol last updated on 08/Jun/20
Commented by Hassen_Timol last updated on 08/Jun/20
How can I prove these equalities please?
$$\mathrm{How}\:\mathrm{can}\:\mathrm{I}\:\mathrm{prove}\:\mathrm{these}\:\mathrm{equalities}\:\mathrm{please}? \\ $$
Commented by john santu last updated on 08/Jun/20
in generally   (√(a+b+2(√(ab)) )) = (√a) + (√b)   (√(a+b−2(√(ab)) )) = (√a) −(√b) ; a > b
$$\mathrm{in}\:\mathrm{generally}\: \\ $$$$\sqrt{\mathrm{a}+\mathrm{b}+\mathrm{2}\sqrt{\mathrm{ab}}\:}\:=\:\sqrt{\mathrm{a}}\:+\:\sqrt{\mathrm{b}}\: \\ $$$$\sqrt{\mathrm{a}+\mathrm{b}−\mathrm{2}\sqrt{\mathrm{ab}}\:}\:=\:\sqrt{\mathrm{a}}\:−\sqrt{\mathrm{b}}\:;\:\mathrm{a}\:>\:\mathrm{b}\: \\ $$
Answered by Aziztisffola last updated on 08/Jun/20
4+2(√3)=1+2×1×(√3)+((√3))^2 =(1+(√3))^2   (√(4+2(√3)))=(√((1+(√3))^2 ))=1+(√3)
$$\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}=\mathrm{1}+\mathrm{2}×\mathrm{1}×\sqrt{\mathrm{3}}+\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} =\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\sqrt{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}=\sqrt{\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }=\mathrm{1}+\sqrt{\mathrm{3}} \\ $$
Commented by Hassen_Timol last updated on 08/Jun/20
Thanks a lot. I didn′t think about that...
$$\mathrm{Thanks}\:\mathrm{a}\:\mathrm{lot}.\:\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{think}\:\mathrm{about}\:\mathrm{that}… \\ $$
Answered by Aziztisffola last updated on 08/Jun/20
((√(3+(√5)))−(√(3−(√5))))^2    =3+(√5)+3−(√5)−2(√((3+(√5))(3−(√5))))   =6−2(√(9−5))=6−2×2=2  (√(3+(√5)))−(√(3−(√5)))=(√2)
$$\left(\sqrt{\mathrm{3}+\sqrt{\mathrm{5}}}−\sqrt{\mathrm{3}−\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} \\ $$$$\:=\mathrm{3}+\sqrt{\mathrm{5}}+\mathrm{3}−\sqrt{\mathrm{5}}−\mathrm{2}\sqrt{\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)} \\ $$$$\:=\mathrm{6}−\mathrm{2}\sqrt{\mathrm{9}−\mathrm{5}}=\mathrm{6}−\mathrm{2}×\mathrm{2}=\mathrm{2} \\ $$$$\sqrt{\mathrm{3}+\sqrt{\mathrm{5}}}−\sqrt{\mathrm{3}−\sqrt{\mathrm{5}}}=\sqrt{\mathrm{2}} \\ $$$$ \\ $$

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