Giải Bài 6.4 trang 9 SGK Toán 11 Kết nối tri thức tập 2 - KNTT

Phương pháp giải

HS sử dụng các tính chất về số mũ:

\(\begin{array}{l} {a^m}.{a^n} = {a^{m + n}};{\quad\quad\quad}\frac{{{a^m}}}{{{a^n}}} = {a^{m - n}};\\ {\left( {{a^m}} \right)^n} = {a^{mn}};{\quad\quad\quad}{\left( {ab} \right)^m} = {a^m}{b^m};\\ {\left( {\frac{a}{b}} \right)^m} = \frac{{{a^m}}}{{{b^m}}}. \end{array}\)

Lời giải chi tiết

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

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

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

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

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

-- Mod Toán 11 HỌC247