Question

Să se calculeze valoarea expresiei


$\it E=\sqrt{x+2\sqrt{x-1}} +\sqrt{x-2\sqrt{x-1}}$

$\it pentru\ \ x\in[1,\ 2]$


Mulțumesc !!!

Answer 1

$\it E = \sqrt{x+2\sqrt{x-1}} +\sqrt{x-2\sqrt{x-1}}\\ \\ E^2 = x+2\sqrt{x-1}+x-2\sqrt{x-1}+2\sqrt{x^2-4(x-1)} \\ \\ E^2 = 2x+2\sqrt{(x-2)^2} = 2x+2|x-2| \\ \\ \text{\it Dar }x\in [1,\ 2] \Rightarrow x-2 \leq 0 \Rightarrow |x-2| = -(x-2) \\ \\ E^2 = 2x-2(x-2) = 2x-2x+4 = 4 \\ E\geq 0 \\ \\ \Rightarrow \boxed{E = 2}$

Sau

$\it E=\sqrt{x+2\sqrt{x-1}} +\sqrt{x-2\sqrt{x-1}}\\ \\ \sqrt{x+2\sqrt{x-1}}= \sqrt a+\sqrt b \\ \\ x+2\sqrt{x-1} = a+b+2\sqrt{ab} \\ \\ a+b = x\\ ab = x-1 \\ \\ t^2-xt+(x-1)=0 \\ \Delta = x^2-4(x-1) = x^2-4x+4 = (x-2)^2 \\ \\ t_{1,2} = \dfrac{x\pm(x-2)}{2} \Rightarrow t_1 = 1,\quad t_2 = \dfrac{2x-2}{2} = x-1 \\ \\ \Rightarrow \sqrt{x+2\sqrt{x-1}} = \sqrt{(\sqrt{x-1}+1)^2 } = |\sqrt{x-1}+1|$

$\it\text{\it Analog: }\sqrt{x-2\sqrt{x-1}} = \sqrt{(\sqrt{x-1}-1)^2} = |\sqrt{x-1}-1|}\\ \\ \Rightarrow E = |\sqrt{x-1}+1|+|\sqrt{x-1}-1| \\ \\ \text{\it Dar }x\in [1,\ 2] \Rightarrow \sqrt{x-1}-1 \leq 0 \\ \\\Rightarrow E = \sqrt{x-1}+1-(\sqrt{x-1}-1 ) \\ \\ \Rightarrow \boxed{E = 2}$