Se utilizează regulile de calcul cu puteri:
[tex]9^{\dfrac{1}{2} } = (3^{2})^{\dfrac{1}{2} } = 3^{2\cdot\dfrac{1}{2} } = 3^{1} = \bf3[/tex]
[tex]49^{-\dfrac{1}{2} } = (7^{2})^{-\dfrac{1}{2} } = 7^{-2\cdot\dfrac{1}{2} } = 7^{-1} = \bf\dfrac{1}{7}[/tex]
[tex]64^{\dfrac{3}{4} } = (2^{6})^{\dfrac{3}{4} } = 2^{6\cdot\dfrac{3}{4} } = 2^{\dfrac{9}{2} } = 2^{4+\dfrac{1}{2} }=2^{4}\cdot2^{\dfrac{1}{2} } = \bf16\sqrt{2}[/tex]
[tex]\bigg(\dfrac{8}{27} \bigg)^{-\dfrac{2}{3} } = \bigg(\dfrac{27}{8} \bigg)^{\dfrac{2}{3} } = \bigg(\dfrac{3^{3} }{2^{3} } \bigg)^{\dfrac{2}{3} } = \bigg[\bigg(\dfrac{3}{2} \bigg)^{3}\bigg]^{\dfrac{2}{3} } = \bigg(\dfrac{3}{2} \bigg)^{3\cdot\dfrac{2}{3} } = \bigg(\dfrac{3}{2} \bigg)^{2} = \dfrac{3^{2} }{2^{2}} = \bf\dfrac{9}{4}[/tex]
[tex]\bigg(\dfrac{16}{49} \bigg)^{-1,5} = \bigg(\dfrac{49}{16} \bigg)^{\dfrac{3}{2} } = \bigg(\dfrac{7^{2} }{4^{2} } \bigg)^{\dfrac{3}{2} } = \bigg[\bigg(\dfrac{7}{4} \bigg)^{2}\bigg]^{\dfrac{3}{2} } = \bigg(\dfrac{7}{4} \bigg)^{2\cdot\dfrac{3}{2} } = \bigg(\dfrac{7}{4} \bigg)^{3} = \bf\dfrac{343}{64}[/tex]
[tex](0,008)^{\dfrac{2}{3} } = \bigg(\dfrac{8}{1000} \bigg)^{\dfrac{2}{3} } = \bigg(\dfrac{1}{125} \bigg)^{\dfrac{2}{3} } = \bigg(\dfrac{1}{5^{3} } \bigg)^{\dfrac{2}{3} } = \bigg[\bigg(\dfrac{1}{5} \bigg)^{3}\bigg]^{\dfrac{2}{3} } = \bigg(\dfrac{1}{5} \bigg)^{3\cdot\dfrac{2}{3} } = \bigg(\dfrac{1}{5} \bigg)^{2} =\bf\dfrac{1}{25}[/tex]
[tex](2,25)^{-\dfrac{3}{2} } = \bigg(\dfrac{225}{100} \bigg)^{-\dfrac{3}{2} } = \bigg(\dfrac{15^{2} }{10^{2} } \bigg)^{-\dfrac{3}{2} } = \bigg[\bigg(\dfrac{15}{10} \bigg)^{2}\bigg]^{-\dfrac{3}{2} } = \bigg(\dfrac{15}{10} \bigg)^{-2\cdot\dfrac{3}{2} } = \bigg(\dfrac{3}{2} \bigg)^{-3} = \bigg(\dfrac{2}{3} \bigg)^{3} =\bf\dfrac{8}{27}[/tex]
______
Reguli de calcul cu puteri:
[tex]\boxed{\boldsymbol{a^{m} \cdot a^{n} = a^{m + n}}; \ \ \ \boldsymbol{a^{m} : a^{n} = a^{m - n}}; \ \ \ \boldsymbol{(a^{m})^{n} = a^{m \cdot n}}; \ \ \boldsymbol{a^{m} \cdot b^{m} = (a \cdot b)^{m}}}[/tex]
[tex]\boxed{\boldsymbol{a^{-m}=\dfrac{1}{a^{m} }}}[/tex]
Show life that you have a thousand reasons to smile
© Copyright 2024 DOKU.TIPS - All rights reserved.
Verified answer
Se utilizează regulile de calcul cu puteri:
[tex]9^{\dfrac{1}{2} } = (3^{2})^{\dfrac{1}{2} } = 3^{2\cdot\dfrac{1}{2} } = 3^{1} = \bf3[/tex]
[tex]49^{-\dfrac{1}{2} } = (7^{2})^{-\dfrac{1}{2} } = 7^{-2\cdot\dfrac{1}{2} } = 7^{-1} = \bf\dfrac{1}{7}[/tex]
[tex]64^{\dfrac{3}{4} } = (2^{6})^{\dfrac{3}{4} } = 2^{6\cdot\dfrac{3}{4} } = 2^{\dfrac{9}{2} } = 2^{4+\dfrac{1}{2} }=2^{4}\cdot2^{\dfrac{1}{2} } = \bf16\sqrt{2}[/tex]
[tex]\bigg(\dfrac{8}{27} \bigg)^{-\dfrac{2}{3} } = \bigg(\dfrac{27}{8} \bigg)^{\dfrac{2}{3} } = \bigg(\dfrac{3^{3} }{2^{3} } \bigg)^{\dfrac{2}{3} } = \bigg[\bigg(\dfrac{3}{2} \bigg)^{3}\bigg]^{\dfrac{2}{3} } = \bigg(\dfrac{3}{2} \bigg)^{3\cdot\dfrac{2}{3} } = \bigg(\dfrac{3}{2} \bigg)^{2} = \dfrac{3^{2} }{2^{2}} = \bf\dfrac{9}{4}[/tex]
[tex]\bigg(\dfrac{16}{49} \bigg)^{-1,5} = \bigg(\dfrac{49}{16} \bigg)^{\dfrac{3}{2} } = \bigg(\dfrac{7^{2} }{4^{2} } \bigg)^{\dfrac{3}{2} } = \bigg[\bigg(\dfrac{7}{4} \bigg)^{2}\bigg]^{\dfrac{3}{2} } = \bigg(\dfrac{7}{4} \bigg)^{2\cdot\dfrac{3}{2} } = \bigg(\dfrac{7}{4} \bigg)^{3} = \bf\dfrac{343}{64}[/tex]
[tex](0,008)^{\dfrac{2}{3} } = \bigg(\dfrac{8}{1000} \bigg)^{\dfrac{2}{3} } = \bigg(\dfrac{1}{125} \bigg)^{\dfrac{2}{3} } = \bigg(\dfrac{1}{5^{3} } \bigg)^{\dfrac{2}{3} } = \bigg[\bigg(\dfrac{1}{5} \bigg)^{3}\bigg]^{\dfrac{2}{3} } = \bigg(\dfrac{1}{5} \bigg)^{3\cdot\dfrac{2}{3} } = \bigg(\dfrac{1}{5} \bigg)^{2} =\bf\dfrac{1}{25}[/tex]
[tex](2,25)^{-\dfrac{3}{2} } = \bigg(\dfrac{225}{100} \bigg)^{-\dfrac{3}{2} } = \bigg(\dfrac{15^{2} }{10^{2} } \bigg)^{-\dfrac{3}{2} } = \bigg[\bigg(\dfrac{15}{10} \bigg)^{2}\bigg]^{-\dfrac{3}{2} } = \bigg(\dfrac{15}{10} \bigg)^{-2\cdot\dfrac{3}{2} } = \bigg(\dfrac{3}{2} \bigg)^{-3} = \bigg(\dfrac{2}{3} \bigg)^{3} =\bf\dfrac{8}{27}[/tex]
______
Reguli de calcul cu puteri:
[tex]\boxed{\boldsymbol{a^{m} \cdot a^{n} = a^{m + n}}; \ \ \ \boldsymbol{a^{m} : a^{n} = a^{m - n}}; \ \ \ \boldsymbol{(a^{m})^{n} = a^{m \cdot n}}; \ \ \boldsymbol{a^{m} \cdot b^{m} = (a \cdot b)^{m}}}[/tex]
[tex]\boxed{\boldsymbol{a^{-m}=\dfrac{1}{a^{m} }}}[/tex]