Đáp án:
$\sqrt[4]{\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}}-\sqrt[4]{\dfrac{3-2\sqrt{2}}{3+2\sqrt{2}}}=2$
Giải thích các bước giải:
$\sqrt[4]{\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}}-\sqrt[4]{\dfrac{3-2\sqrt{2}}{3+2\sqrt{2}}}$
$=\dfrac{\sqrt[4]{2+2\sqrt{2}+1}}{\sqrt[4]{2-2\sqrt{2}+1}}-\dfrac{\sqrt[4]{2-2\sqrt{2}+1}}{\sqrt[4]{2+2\sqrt{2}+1}}$
$=\dfrac{\sqrt[4]{(\sqrt{2}+1)^2}}{\sqrt[4]{(\sqrt{2}-1)^2}}-\dfrac{\sqrt[4]{(\sqrt{2}-1)^2}}{\sqrt[4]{(\sqrt{2}+1)^2}}$
$=\dfrac{\sqrt{\sqrt{2}+1}}{\sqrt{\sqrt{2}-1}}-\dfrac{\sqrt{\sqrt{2}-1}}{\sqrt{\sqrt{2}+1}}$
$=\dfrac{(\sqrt{\sqrt{2}+1})^2-(\sqrt{\sqrt{2}-1})^2}{(\sqrt{\sqrt{2}-1})(\sqrt{\sqrt{2}+1})}$
$=\dfrac{\sqrt{2}+1-\sqrt{2}+1}{\sqrt{(\sqrt{2}-1)(\sqrt{2}+1)}}$
$=\dfrac{2}{\sqrt{2-1}}$
$=\dfrac{2}{\sqrt{1}}$
$=\dfrac{2}{1}$
$=2$
Chúc bạn học tốt !!!
