Question

\(\dfrac{\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}}{\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}}\) làm giúp mik bài này nha mọi người,mik đang cần gấp,mik cảm ơn nha!!!!

Answer 1

\(\dfrac{\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}}{\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}}\)

\(=\dfrac{\sqrt{1+2\sqrt{2}+2}+\sqrt{2-2\sqrt{2}+1}}{\sqrt{1+2\sqrt{2}+2}-\sqrt{2-2\sqrt{2}+1}}\)

\(=\dfrac{\sqrt{\left(1+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(1+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}}\)

\(=\dfrac{\left|1+\sqrt{2}\right|+\left|\sqrt{2}-1\right|}{\left|1+\sqrt{2}\right|-\left|\sqrt{2}-1\right|}\)

\(=\dfrac{1+\sqrt{2}+\sqrt{2}-1}{1+\sqrt{2}-\left(\sqrt{2}-1\right)}\)

\(=\dfrac{2\sqrt{2}}{1+\sqrt{2}-\sqrt{2}+1}\)

\(=\dfrac{2\sqrt{2}}{2}=\sqrt{2}\)

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