Giải Bài tập 6.4 trang 9 sgk Toán 11 tập 2 Kết nối

a) $A=\frac{(\mathrm{x}^{\frac{1}{3}}\sqrt{\mathrm{y}}+\mathrm{y}\frac{1}{3})(\sqrt[6]{\mathrm{x}}-\sqrt[6]{\mathrm{y}})}{(\sqrt[6]{\mathrm{x}}+\sqrt[6]{\mathrm{y}})(\sqrt[6]{\mathrm{x}}-\sqrt[6]{\mathrm{y}})}$

$A=\frac{\mathrm{x}^{\frac{2}{6}}\mathrm{y}^{\frac{1}{2}}-\mathrm{x}^{\frac{1}{6}}\mathrm{y}^{\frac{1}{3}}+\mathrm{y}\frac{1}{3}\sqrt[6]{\mathrm{x}}-\mathrm{y}\frac{1}{3}\sqrt[6]{\mathrm{y}}}{\mathrm{x}^{\frac{1}{6}}+\mathrm{y}^{\frac{1}{6}}}$

$A=\frac{\mathrm{x}^{\frac{1}{6}}\mathrm{y}^{\frac{1}{2}}-\mathrm{x}^{\frac{1}{6}}\mathrm{y}^{\frac{1}{3}}+\mathrm{y}\frac{1}{3}\sqrt[6]{\mathrm{x}}-\mathrm{y}\frac{1}{3}\sqrt[6]{\mathrm{y}}}{1}$

$A={\mathrm{x}^{\frac{1}{6}}\mathrm{y}^{\frac{1}{3}}-\frac{1}{3}(\sqrt[6]{\mathrm{y}}-\sqrt[6]{\mathrm{x}})}$

b) $B=(\frac{\mathrm{x}^{\sqrt{3}}}{y^{\sqrt{3}-1}})^{\sqrt{3}+1}\cdot \frac{x^{-\sqrt{3}-1}}{y^{-2}}$

$B=(\frac{\mathrm{x}^{\sqrt{3}(\sqrt{3}+1)}}{y^{\sqrt{3}-1}(\sqrt{3}+1)})\cdot \frac{x^{-\sqrt{3}-1}}{y^{-2}}$

$B=\frac{\mathrm{x}^{3}}{y^{\sqrt{3}+1}}\cdot \frac{x^{-\sqrt{3}-1}}{y^{-2}}$

$B=\frac{\mathrm{x}^{3-\sqrt{3}-1}}{y^{\sqrt{3}+1-(-2)}}$

$B={\frac{\mathrm{x}^{2-\sqrt{3}}}{y^{\sqrt{3}+3}}}$