Giải SBT Toán học 11 CTST Bài 1: Phép tính lũy thừa có đáp án

a) ${(\frac{1}{\sqrt[3]{5}})}^{0}=1$

b) ${(\frac{2}{5})}^{-2}={(\frac{5}{2})}^{2}=\frac{{5}^{2}}{{2}^{2}}=\frac{25}{4}$

c) ${(-\frac{1}{3})}^{-\ 4}=\frac{1}{{(\frac{1}{3})}^{4}}=\frac{1}{\frac{1}{81}}=81$

d) (−55)$^{0}$ = 1.

e) ${2}^{-8}.\ {2}^{5}={2}^{-8+5}={2}^{-3}={(\frac{1}{2})}^{3}=\frac{1}{{2}^{3}}=\frac{1}{8}$

g) $\frac{{3}^{4}}{{({3}^{-2})}^{-3}}=\frac{{3}^{4}}{{{3}^{(-2)}}^{.(-3)}}=\frac{{3}^{4}}{{3}^{6}}$$={3}^{4-6}={3}^{-2}=\frac{1}{{3}^{2}}=\frac{1}{9}$

a) $\sqrt[3]{0{,}001}=\sqrt[3]{{(0{,}1)}^{3}}=0{,}1$ .

b) $\sqrt[5]{-32}=\sqrt[5]{(-1).32}=\sqrt[5]{{(-1)}^{5}{.2}^{5}}$

$=\sqrt[5]{{(-1)}^{5}}\ .\ \sqrt[5]{{2}^{5}}=(-1)\ .\ 2=-2$

c) $\sqrt[4]{\frac{81}{16}}=\sqrt[4]{\frac{{3}^{4}}{{2}^{4}}}=\sqrt[4]{{(\frac{3}{2})}^{4}}=\frac{3}{2}$

d) $-\sqrt[6]{{100}^{3}}=-\sqrt[6]{{({10}^{2})}^{3}}=-\sqrt[6]{{10}^{6}}=-10$

e) $\sqrt[4]{{(\sqrt{3}-2)}^{4}}=|\sqrt{3}-2|=2-\sqrt{3}$

g) $\sqrt[5]{{(2-\sqrt{5})}^{5}}=2-\sqrt{5}$

a) $\sqrt[4]{125}.\sqrt[4]{5}=\sqrt[4]{125.5}=\sqrt[4]{{5}^{3}.5}=\sqrt[4]{{5}^{4}}=5$ .

b) $\frac{\sqrt[4]{243}}{\sqrt[4]{3}}=\sqrt[4]{\frac{243}{3}}=\sqrt[4]{81}=\sqrt[4]{{3}^{4}}=3$ .

c) $\frac{\sqrt[3]{3}}{\sqrt[3]{24}}=\sqrt[3]{\frac{3}{24}}=\sqrt[3]{\frac{1}{8}}=\sqrt[3]{{(\frac{1}{2})}^{3}}=\frac{1}{2}$ .

d) $\sqrt{\sqrt[3]{64}}=\sqrt[6]{64}=\sqrt[6]{{2}^{6}}=2$ .

e) $\sqrt[4]{3\sqrt[3]{3}}=\sqrt[4]{{(\sqrt[3]{3})}^{3}.\sqrt[3]{3}}=\sqrt[4]{{(\sqrt[3]{3})}^{4}}=\sqrt[3]{3}$ .

g) ${(-\sqrt[6]{4})}^{3}={(-1)}^{3}.{(\sqrt[6]{4})}^{3}=-\sqrt[6]{{4}^{3}}$

$=-\sqrt[6]{{({2}^{2})}^{3}}=-\sqrt[6]{{2}^{6}}=-2$

a) $\sqrt[3]{135}-5\sqrt[3]{5}=\sqrt[3]{{3}^{3}\ .\ 5}-5\sqrt[3]{5}=\sqrt[3]{{3}^{3}}\ .\ \sqrt[3]{5}-5\sqrt[3]{5}$

$=3\sqrt[3]{5}-5\sqrt[3]{5}=-2\sqrt[3]{5}$

b) $\sqrt[4]{\sqrt[3]{81}}+3\sqrt[3]{3}=\sqrt[4]{\sqrt[3]{{3}^{4}}}+3\sqrt[3]{3}=\sqrt[3]{3}+3\sqrt[3]{3}=4\sqrt[3]{3}$

c) $\sqrt[4]{\sqrt[5]{16}}+\sqrt[5]{64}+2\sqrt[5]{2}=\sqrt[4]{\sqrt[5]{{2}^{4}}}+\sqrt[5]{{2}^{6}}+2\sqrt[5]{2}$

$=\sqrt[5]{2}+2\sqrt[5]{2}+2\sqrt[5]{2}=5\sqrt[5]{2}$

d) ${(\sqrt[4]{5})}^{5}-\sqrt{\sqrt[4]{25}}={(\sqrt[4]{5})}^{4}\ .\ \sqrt[4]{5}-\sqrt{\sqrt[4]{{5}^{2}}}$

$=5\sqrt[4]{5}-\sqrt[4]{5}=4\sqrt[4]{5}$

a) ${8}^{-\frac{2}{3}}={({2}^{3})}^{-\frac{2}{3}}={2}^{-2}=\frac{1}{{2}^{2}}=\frac{1}{4}$

b) ${32}^{-\frac{2}{5}}={(\frac{1}{32})}^{\frac{2}{5}}=\frac{1}{\sqrt[5]{{32}^{2}}}=\frac{1}{\sqrt[5]{{({2}^{5})}^{2}}}=\frac{1}{{2}^{2}}=\frac{1}{4}$

c) ${81}^{1{,}25}={81}^{\frac{5}{4}}=\sqrt[4]{{81}^{5}}=\sqrt[4]{{({3}^{4})}^{5}}={3}^{5}=243$

d) ${1000}^{-\frac{5}{3}}={(\frac{1}{1000})}^{\frac{5}{3}}=\frac{1}{\sqrt[3]{{1000}^{5}}}=\frac{1}{\sqrt[3]{{({10}^{3})}^{5}}}=\frac{1}{{10}^{5}}={10}^{-5}$

e) ${(\frac{16}{81})}^{-\frac{1}{4}}={(\frac{81}{16})}^{\frac{1}{4}}=\sqrt[4]{\frac{81}{16}}=\sqrt[4]{\frac{{3}^{4}}{{2}^{4}}}=\sqrt[4]{{(\frac{3}{2})}^{4}}=\frac{3}{2}$

g) ${(\frac{8}{27})}^{-\frac{2}{3}}={(\frac{27}{8})}^{\frac{2}{3}}=\sqrt[3]{{(\frac{27}{8})}^{2}}$

$=\sqrt[3]{{[{(\frac{3}{2})}^{3}]}^{2}}={(\frac{3}{2})}^{2}=\frac{9}{4}$