Bài 18 trang 81 SGK Đại số và Giải tích 12 Nâng cao

Viết các biểu thức sau dưới dạng lũy thừa của một số với số mũ hữu tỉ:

LG a

$\sqrt[4]{x^2\sqrt[3]{x}}\left(x>0\right);$       

Phương pháp giải:

Lưu ý: $\sqrt[n]{a}=a^{\frac{1}{n}}(a>0)$; $a^m.a^n=a^{m+n}$

Lời giải chi tiết:

$\sqrt[4]{x^2\sqrt[3]{x}}={\left(x^2.x^{\frac{1}{3}}\right)}^{\frac{1}{4}}={\left(x^{\frac{7}{3}}\right)}^{\frac{1}{4}}=x^{\frac{7}{12}}$

Cách trình bày khác:

$\begin{array}{l}\sqrt[4]{x^2\sqrt[3]{x}}=\sqrt[4]{x^2.x^{\frac{1}{3}}}=\sqrt[4]{x^{\frac{7}{3}}}\\ ={\left(x^{\frac{7}{3}}\right)}^{\frac{1}{4}}=x^{\frac{7}{3}.\frac{1}{4}}=x^{\frac{7}{12}}\end{array}$

LG b

$\sqrt[5]{\frac{b}{a}\sqrt[3]{\frac{a}{b}}}\left(a>0,b>0\right);$

Lời giải chi tiết:

$\sqrt[5]{\frac{b}{a}\sqrt[3]{\frac{a}{b}}}={\left(\frac{b}{a}{\left(\frac{a}{b}\right)}^{\frac{1}{3}}\right)}^{\frac{1}{5}}$

$={\left({\left(\frac{a}{b}\right)}^{-1}{\left(\frac{a}{b}\right)}^{\frac{1}{3}}\right)}^{\frac{1}{5}}={\left({\left(\frac{a}{b}\right)}^{-\frac{2}{3}}\right)}^{\frac{1}{5}}$

$={\left(\frac{a}{b}\right)}^{-\frac{2}{15}}$

Cách trình bày khác:

$\begin{array}{l}\sqrt[5]{\frac{b}{a}.\sqrt[3]{\frac{a}{b}}}=\sqrt[5]{{\left(\frac{a}{b}\right)}^{-1}.{\left(\frac{a}{b}\right)}^{\frac{1}{3}}}\\ =\sqrt[5]{{\left(\frac{a}{b}\right)}^{-1}.{\left(\frac{a}{b}\right)}^{\frac{1}{3}}}=\sqrt[5]{{\left(\frac{a}{b}\right)}^{-\frac{2}{3}}}\\ ={\left[{\left(\frac{a}{b}\right)}^{-\frac{2}{3}}\right]}^{\frac{1}{5}}={\left(\frac{a}{b}\right)}^{-\frac{2}{3}.\frac{1}{5}}\\ ={\left(\frac{a}{b}\right)}^{-\frac{2}{15}}\end{array}$

LG c

$\sqrt[3]{\frac{2}{3}\sqrt[3]{\frac{2}{3}}\sqrt{\frac{2}{3}}};$ 

Lời giải chi tiết:

$\begin{array}{l}\sqrt[3]{\frac{2}{3}\sqrt[3]{\frac{2}{3}\sqrt{\frac{2}{3}}}}=\sqrt[3]{\frac{2}{3}\sqrt[3]{\frac{2}{3}.{\left(\frac{2}{3}\right)}^{\frac{1}{2}}}}\\ =\sqrt[3]{\frac{2}{3}\sqrt[3]{{\left(\frac{2}{3}\right)}^{\frac{3}{2}}}}=\sqrt[3]{\frac{2}{3}{\left[{\left(\frac{2}{3}\right)}^{\frac{3}{2}}\right]}^{\frac{1}{3}}}\\ =\sqrt[3]{\frac{2}{3}.{\left(\frac{2}{3}\right)}^{\frac{3}{2}.\frac{1}{3}}}=\sqrt[3]{\frac{2}{3}.{\left(\frac{2}{3}\right)}^{\frac{1}{2}}}\\ =\sqrt[3]{{\left(\frac{2}{3}\right)}^{1+\frac{1}{2}}}=\sqrt[3]{{\left(\frac{2}{3}\right)}^{\frac{3}{2}}}\\ ={\left[{\left(\frac{2}{3}\right)}^{\frac{3}{2}}\right]}^{\frac{1}{3}}={\left(\frac{2}{3}\right)}^{\frac{3}{2}.\frac{1}{3}}\\ ={\left(\frac{2}{3}\right)}^{\frac{1}{2}}\end{array}$

LG d

$\sqrt{a\sqrt{a\sqrt{a\sqrt{a}}}}:a^{\frac{11}{16}}\left(a>0\right).$

Lời giải chi tiết:

$\begin{array}{l}\sqrt{a\sqrt{a\sqrt{a\sqrt{a}}}}:a^{\frac{11}{16}}\\ =\sqrt{a\sqrt{a\sqrt{a.a^{\frac{1}{2}}}}}:a^{\frac{11}{16}}\\ =\sqrt{a\sqrt{a\sqrt{a^{\frac{3}{2}}}}}:a^{\frac{11}{16}}\\ =\sqrt{a\sqrt{a{\left(a^{\frac{3}{2}}\right)}^{\frac{1}{2}}}}:a^{\frac{11}{16}}\\ =\sqrt{a\sqrt{a.a^{\frac{3}{2}.\frac{1}{2}}}}:a^{\frac{11}{16}}\\ =\sqrt{a\sqrt{a.a^{\frac{3}{4}}}}:a^{\frac{11}{16}}\\ =\sqrt{a\sqrt{a^{1+\frac{3}{4}}}}:a^{\frac{11}{16}}\\ =\sqrt{a\sqrt{a^{\frac{7}{4}}}}:a^{\frac{11}{16}}\\ =\sqrt{a.{\left(a^{\frac{7}{4}}\right)}^{\frac{1}{2}}}:a^{\frac{11}{16}}\\ =\sqrt{a.a^{\frac{7}{4}.\frac{1}{2}}}:a^{\frac{11}{16}}\\ =\sqrt{a.a^{\frac{7}{8}}}:a^{\frac{11}{16}}\\ =\sqrt{a^{1+\frac{7}{8}}}:a^{\frac{11}{16}}\\ =\sqrt{a^{\frac{15}{8}}}:a^{\frac{11}{16}}\\ ={\left(a^{\frac{15}{8}}\right)}^{\frac{1}{2}}:a^{\frac{11}{16}}\\ =a^{\frac{15}{8}.\frac{1}{2}}:a^{\frac{11}{16}}\\ =a^{\frac{15}{16}}:a^{\frac{11}{16}}\\ =a^{\frac{15}{16}-\frac{11}{16}}\\ =a^{\frac{1}{4}}\end{array}$

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