Đáp án:
c. $\frac{-1-\sqrt[]{2}}{\sqrt[]{3}+\sqrt[]{2}}$
d. $1 + \sqrt[]{2}$
e. $\sqrt[]{\frac{x}{y}}$
f. $\frac{\sqrt[]{a}-\sqrt[]{b}}{\sqrt[]{ab}-1}$
Giải thích các bước giải:
c. $\frac{2\sqrt[]{15}-2\sqrt[]{10}+\sqrt[]{6}-3}{2\sqrt[]{5}-2\sqrt[]{10}-\sqrt[]{3}+\sqrt[]{6}}$
= $\frac{2\sqrt[]{5×3}-2\sqrt[]{5×2}+\sqrt[]{3×2}-3}{2\sqrt[]{5}-2\sqrt[]{5×2}-\sqrt[]{3}+\sqrt[]{3×2}}$
= $\frac{\sqrt[]{3}(2\sqrt[]{5}-\sqrt[]{3})-\sqrt[]{2}(2\sqrt[]{5}-\sqrt[]{3})}{2\sqrt[]{5}(1-\sqrt[]{5})-\sqrt[]{3}(1-\sqrt[]{2})}$
= $\frac{(2\sqrt[]{5}-\sqrt[]{3})(\sqrt[]{3}-\sqrt[]{2})}{(1-\sqrt[]{2})(2\sqrt[]{5}-\sqrt[]{3})}$
= $\frac{\sqrt[]{3}-\sqrt[]{2}}{1-\sqrt[]{2}}$
= $\frac{(\sqrt[]{3}-\sqrt[]{2})(\sqrt[]{3}+\sqrt[]{2})(1+\sqrt[]{2})}{(1-\sqrt[]{2})(1+\sqrt[]{2})(\sqrt[]{3}+\sqrt[]{2})}$
= $\frac{(3-2)(1+\sqrt[]{2})}{(1-2)(\sqrt[]{3}+\sqrt[]{2})}$
= $\frac{-1-\sqrt[]{2}}{\sqrt[]{3}+\sqrt[]{2}}$
f. $\frac{\sqrt[]{a}+a\sqrt[]{b}-\sqrt[]{b}-b\sqrt[]{a}}{ab-1}$
= $\frac{\sqrt[]{a}(1+\sqrt[]{ab})-\sqrt[]{b}(1+\sqrt[]{ab})}{(\sqrt[]{ab}-1)(\sqrt[]{ab}+1)}$
= $\frac{(1+\sqrt[]{ab})(\sqrt[]{a}-\sqrt[]{b})}{(\sqrt[]{ab}-1)(\sqrt[]{ab}+1)}$
= $\frac{\sqrt[]{a}-\sqrt[]{b}}{\sqrt[]{ab}-1}$
e. $\frac{x+\sqrt[]{xy}}{y+\sqrt[]{xy}}$
= $\frac{\sqrt[]{x}(\sqrt[]{x}+\sqrt[]{y})}{\sqrt[]{y}(\sqrt[]{y}+\sqrt[]{x})}$
= $\sqrt[]{\frac{x}{y}}$
d. $\frac{\sqrt[]{2}+\sqrt[]{3}+\sqrt[]{6}+\sqrt[]{8}+\sqrt[]{16}}{\sqrt[]{2}+\sqrt[]{3}+\sqrt[]{4}}$
= $\frac{\sqrt[]{2}+\sqrt[]{3}+\sqrt[]{6}+2\sqrt[]{2}+4}{\sqrt[]{2}+\sqrt[]{3}+2}$
= $\frac{\sqrt[]{2}(1+\sqrt[]{2})+\sqrt[]{3}(1+\sqrt[]{2})+2(\sqrt[]{2}+1)}{\sqrt[]{2}+\sqrt[]{3}+2}$
= $\frac{(1+\sqrt[]{2})(\sqrt[]{2}+\sqrt[]{3}+2)}{\sqrt[]{2}+\sqrt[]{3}+2}$
= $1 + \sqrt[]{2}$
