Question Number 149876 by liberty last updated on 08/Aug/21
Answered by MJS_new last updated on 08/Aug/21
$${y}\geqslant\mathrm{0} \\ $$$${y}=\sqrt{{x}+\mathrm{2}\sqrt{{x}−\mathrm{1}}}+\sqrt{{x}−\mathrm{2}\sqrt{{x}−\mathrm{1}}} \\ $$$$\mathrm{squaring} \\ $$$${y}^{\mathrm{2}} =\left({x}+\mathrm{2}\sqrt{{x}−\mathrm{1}}\right)+\mathrm{2}\sqrt{\left({x}+\mathrm{2}\sqrt{{x}−\mathrm{1}}\right)\left({x}−\mathrm{2}\sqrt{{x}−\mathrm{1}}\right)}+\left({x}−\mathrm{2}\sqrt{{x}−\mathrm{1}}\right) \\ $$$${y}^{\mathrm{2}} =\mathrm{2}{x}+\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{4}\left({x}−\mathrm{1}\right)} \\ $$$${y}^{\mathrm{2}} =\mathrm{2}{x}+\mathrm{2}\sqrt{\left({x}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$$${y}^{\mathrm{2}} =\mathrm{2}{x}+\mathrm{2}\mid{x}−\mathrm{2}\mid \\ $$$${x}\leqslant\mathrm{2}\:\Rightarrow\:\mid{x}−\mathrm{2}\mid=\mathrm{2}−{x} \\ $$$${y}^{\mathrm{2}} =\mathrm{2}{x}+\mathrm{2}\left(\mathrm{2}−{x}\right) \\ $$$${y}^{\mathrm{2}} =\mathrm{4} \\ $$$${y}>\mathrm{0}\:\Rightarrow\:{y}=\mathrm{2} \\ $$
Answered by john_santu last updated on 08/Aug/21
![method 2 let 2(√(x−1)) = u ; x−1=(1/4)u^2 ⇒x=((u^2 +4)/4) ; 1≤((u^2 +4)/4)≤2 ⇒4≤u^2 +4≤8 ; 0≤u^2 ≤4 ⇒0≤u≤2 the equality become to (√(((u^2 +4)/4)+u)) +(√(((u^2 +4)/4)−u)) = (1/2) [(√(u^2 +4u+4)) +(√(u^2 −4u+4)) ]= (1/2)[ ∣u+2∣ +∣u−2∣ ] = (1/2)[ u+2−(u−2)] =(1/2)×4=2](https://www.tinkutara.com/question/Q149878.png)
$${method}\:\mathrm{2} \\ $$$${let}\:\mathrm{2}\sqrt{{x}−\mathrm{1}}\:=\:{u}\:;\:{x}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{4}}{u}^{\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{{u}^{\mathrm{2}} +\mathrm{4}}{\mathrm{4}}\:;\:\mathrm{1}\leqslant\frac{{u}^{\mathrm{2}} +\mathrm{4}}{\mathrm{4}}\leqslant\mathrm{2} \\ $$$$\Rightarrow\mathrm{4}\leqslant{u}^{\mathrm{2}} +\mathrm{4}\leqslant\mathrm{8}\:;\:\mathrm{0}\leqslant{u}^{\mathrm{2}} \leqslant\mathrm{4} \\ $$$$\Rightarrow\mathrm{0}\leqslant{u}\leqslant\mathrm{2} \\ $$$${the}\:{equality}\:{become}\:{to} \\ $$$$\sqrt{\frac{{u}^{\mathrm{2}} +\mathrm{4}}{\mathrm{4}}+{u}}\:+\sqrt{\frac{{u}^{\mathrm{2}} +\mathrm{4}}{\mathrm{4}}−{u}}\:= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:\left[\sqrt{{u}^{\mathrm{2}} +\mathrm{4}{u}+\mathrm{4}}\:+\sqrt{{u}^{\mathrm{2}} −\mathrm{4}{u}+\mathrm{4}}\:\right]= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\:\mid{u}+\mathrm{2}\mid\:+\mid{u}−\mathrm{2}\mid\:\right]\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\:{u}+\mathrm{2}−\left({u}−\mathrm{2}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{4}=\mathrm{2} \\ $$