Bài 6.4 trang 9

Đề bài

Cho x, y là các số thực dương. Rút gọn các biểu thức sau:

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

b) $B={\left(\frac{x^{\sqrt{3}}}{y^{\sqrt{3}-1}}\right)}^{\sqrt{3}+1}.\frac{x^{-\sqrt{3}-1}}{y^{-2}}$.

Lời giải chi tiết

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

$=\frac{x^{\frac{1}{3}}.y^{\frac{1}{2}}+y^{\frac{1}{3}}.x^{\frac{1}{2}}}{x^{\frac{1}{6}}+y^{\frac{1}{6}}}$

$=\frac{x^{\frac{1}{3}}.y^{\frac{1}{6}+\frac{1}{3}}+y^{\frac{1}{3}}.x^{\frac{1}{6}+\frac{1}{3}}}{x^{\frac{1}{6}}+y^{\frac{1}{6}}}$

$=\frac{x^{\frac{1}{3}}.y^{\frac{1}{6}}.y^{\frac{1}{3}}+y^{\frac{1}{3}}.x^{\frac{1}{6}}.x^{\frac{1}{3}}}{x^{\frac{1}{6}}+y^{\frac{1}{6}}}$

$=\frac{x^{\frac{1}{3}}.y^{\frac{1}{3}}\left(y^{\frac{1}{6}}+x^{\frac{1}{6}}\right)}{x^{\frac{1}{6}}+y^{\frac{1}{6}}}$

$=\sqrt[3]{3x}.\sqrt[3]{3y}$

$=\sqrt[3]{3xy}$.

b) $B={\left(\frac{x^{\sqrt{3}}}{y^{\sqrt{3}-1}}\right)}^{\sqrt{3}+1}.\frac{x^{-\sqrt{3}-1}}{y^{-2}}$

$=\frac{x^{\sqrt{3}.\left(\sqrt{3}+1\right)}}{y^{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}}.\frac{x^{-\sqrt{3}-1}}{y^{-2}}$

$=\frac{x^{3+\sqrt{3}}}{y^2}.\frac{x^{-\left(\sqrt{3}+1\right)}}{y^{-2}}$

$=\frac{x^{3+\sqrt{3}}}{y^2}.\frac{y^2}{x^{\sqrt{3}+1}}$

$=\frac{x^{3+\sqrt{3}}}{x^{\sqrt{3}+1}}$

$=x^{3+\sqrt{3}-\sqrt{3}-1}$

$=x^2$.