Tính
$\frac{3+\sqrt[]{5}}{\sqrt[]{2}+\sqrt[]{3+\sqrt[]{5}}}$ +$\frac{3-\sqrt[]{5}}{\sqrt[]{2}-\sqrt[]{3-\sqrt[]{5}}}$
Tính
$\frac{3+\sqrt[]{5}}{\sqrt[]{2}+\sqrt[]{3+\sqrt[]{5}}}$ +$\frac{3-\sqrt[]{5}}{\sqrt[]{2}-\sqrt[]{3-\sqrt[]{5}}}$
Đáp án:vote cho mình nha hc tốt
Giải thích các bước giải:
$\frac{3+\sqrt5}{\sqrt2+\sqrt{3+\sqrt5}}$ + $\frac{3-\sqrt5}{\sqrt2-\sqrt{3-\sqrt5}}$
=$\frac{\sqrt2(3+\sqrt5)}{\sqrt2(\sqrt2+\sqrt{3+\sqrt5})}$ + $\frac{\sqrt2(3-\sqrt5)}{\sqrt2(\sqrt2-\sqrt{3-\sqrt5})}$
=$\frac{\sqrt2(3+\sqrt5)}{2+\sqrt{6+2\sqrt5}}$ + $\frac{\sqrt2(3-\sqrt5)}{2-\sqrt{6-2\sqrt5}}$
=$\frac{\sqrt2(3+\sqrt5)}{2+\sqrt{(\sqrt5+1)^2}}$ + $\frac{\sqrt2(3-\sqrt5)}{2-\sqrt{(\sqrt5-1)^2}}$
=$\frac{\sqrt2(3+\sqrt5)}{2+(\sqrt5+1)}$ + $\frac{\sqrt2(3-\sqrt5)}{2-(\sqrt5-1)}$
=$\frac{\sqrt2(3+\sqrt5)}{3+\sqrt5}$ + $\frac{\sqrt2(3-\sqrt5)}{3-\sqrt5}$=$\sqrt[]{2}$ +$\sqrt[]{2}$ = 2$\sqrt[]{2}$