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Question Number 183935 by SulaymonNorboyev last updated on 31/Dec/22
(√(x^2 +2))+(√(x^2 −4))=A  (√(x^2 +2))−(√(x^2 −4))=?
$$\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}={A} \\ $$$$\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}=? \\ $$
Answered by Rasheed.Sindhi last updated on 31/Dec/22
(√(x^2 +2))+(√(x^2 −4))=A    (√(x^2 +2))−(√(x^2 −4))=?  ((√(x^2 +2))+(√(x^2 −4)) )((√(x^2 +2)) −(√(x^2 −4)) )                                     =A((√(x^2 +2)) −(√(x^2 −4)) )  ((√(x^2 +2)) −(√(x^2 −4)) )=(1/A){(x^2 +2)−(x^2 −4)}  (√(x^2 +2)) −(√(x^2 −4)) =(6/A)
$$\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}={A}\: \\ $$$$\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}=? \\ $$$$\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}\:\right)\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}\:−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={A}\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}\:−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}\:\right) \\ $$$$\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}\:−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}\:\right)=\frac{\mathrm{1}}{{A}}\left\{\left({x}^{\mathrm{2}} +\mathrm{2}\right)−\left({x}^{\mathrm{2}} −\mathrm{4}\right)\right\} \\ $$$$\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}\:−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}\:=\frac{\mathrm{6}}{{A}} \\ $$
Commented by SulaymonNorboyev last updated on 31/Dec/22
  THANKS YOU
$$ \\ $$$${THANKS}\:{YOU} \\ $$
Answered by Rasheed.Sindhi last updated on 31/Dec/22
(√(x^2 +2))+(√(x^2 −4))=A  (√(x^2 +2))−(√(x^2 −4))=?  a=(√(x^2 +2)) , b=(√(x^2 −4))   a+b=A.........(i)  a^2 −b^2 =(x^2 +2)−(x^2 −4)=6  a−b=((a^2 −b^2 )/(a+b))=(6/A)  (√(x^2 +2)) −(√(x^2 −4))=(6/A)
$$\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}={A} \\ $$$$\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}=? \\ $$$${a}=\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}\:,\:{b}=\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}\: \\ $$$${a}+{b}={A}………\left({i}\right) \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\left({x}^{\mathrm{2}} +\mathrm{2}\right)−\left({x}^{\mathrm{2}} −\mathrm{4}\right)=\mathrm{6} \\ $$$${a}−{b}=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}+{b}}=\frac{\mathrm{6}}{{A}} \\ $$$$\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}\:−\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}=\frac{\mathrm{6}}{{A}} \\ $$
Answered by manolex last updated on 31/Dec/22
m+n=A      find   m−n  (m+n)(m−n)=(m−n)A  x^2 +2−x^2 +4=(m−n)A  (6/A)=(√(x^2 +2))− (√(x^2 −4))      is answer
$${m}+{n}={A}\:\:\:\:\:\:{find}\:\:\:{m}−{n} \\ $$$$\left({m}+{n}\right)\left({m}−{n}\right)=\left({m}−{n}\right){A} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}−{x}^{\mathrm{2}} +\mathrm{4}=\left({m}−{n}\right){A} \\ $$$$\frac{\mathrm{6}}{{A}}=\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}−\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}\:\:\:\:\:\:{is}\:{answer} \\ $$

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