$* \ A=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}} \ \ \ \ \ \ \ \ (x\geq 2)
\\=\sqrt{x-2+2\sqrt{x-2}.\sqrt{2}+2}+\sqrt{x-2-2\sqrt{x-2}.\sqrt{2}+2}
\\=\sqrt{(\sqrt{x-2}+\sqrt{2})^2}+\sqrt{(\sqrt{x-2}-\sqrt{2})^2}
\\=\sqrt{x-2}+\sqrt{2}+|\sqrt{x-2}-\sqrt{2}|$
Nếu $\sqrt{x-2}-\sqrt{2}\geq 0\Leftrightarrow x\geq 4$ thì $A=\sqrt{x-2}+\sqrt{2}+\sqrt{x-2}-\sqrt{2}=2\sqrt{x-2}$
Nếu $\sqrt{x-2}-\sqrt{2}< 0\Leftrightarrow 2\leq x<4$ thì $A=\sqrt{x-2}+\sqrt{2}+\sqrt{2}-\sqrt{x-2}=2\sqrt{2}$
Ta có:
$\left (1+\dfrac{1}{a}-\dfrac{1}{a+1} \right )^2=1+\dfrac{1}{a^2}+\dfrac{1}{(a+1)^2}+\dfrac{2}{a}-\dfrac{2}{a(a+1)}-\dfrac{2}{a+1}
\\=1+\dfrac{1}{a^2}+\dfrac{1}{(a+1)^2}+\dfrac{2(a+1)-2-2a}{a(a+1)}=1+\dfrac{1}{a^2}+\dfrac{1}{(a+1)^2}
\\\Rightarrow B=\sqrt{1+\dfrac{1}{a^2}+\dfrac{1}{(a+1)^2}}=\sqrt{\left (1+\dfrac{1}{a}-\dfrac{1}{a+1} \right )^2}
\\=\left | 1+\dfrac{1}{a}-\dfrac{1}{a+1} \right |=\left | \dfrac{a(a+1)+a+1-a}{a(a+1)} \right |=\dfrac{|a^2+a+a+1-a|}{a(a+1)}
\\=\dfrac{|a^2+a+1|}{a(a+1)}=\dfrac{a^2+a+1}{a(a+1)}$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
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