Question

Repede va rog! e tema pentru luni!

Answer 1

[tex]\displaystyle a)~~\left( - \frac{1235}{1997} \right)^0=\boxed{1} \\ \\ b)~~\left( - \frac{36}{47} \right)^1= \boxed{- \frac{36}{47}} \\ \\ c)~~\left( - \frac{1997}{1997} \right)^1= (-1)^{1997}= \boxed{- 1} \\ \\ d)~~\left( - \frac{17}{34} \right)^4= \left( \frac{17}{34} \right)^4 = \left( \frac{1}{2} \right)^4 = \frac{1}{2^4} = \boxed{\frac{1}{16} } \\ \\ e)~~\left( -4 \frac{32}{128} \right)^{1997}= \left(-\frac{32}{32} \right)^{1997} =(-1)^{1997}=\boxed{-1}[/tex]


[tex]\displaystyle f)~~0^{2015}= \boxed{0} \\ \\ g)~~ \left( -\frac{23}{23}\right )^{-1} = \left( -\frac{23}{23}\right )^{1} = -\frac{23}{23} =\boxed{-1} \\ \\ h)~~ \left( -1,5\right )^{-2} = \left( -\frac{15}{10}\right )^{-2} = \left( -\frac{3}{2}\right )^{-2}=\left( -\frac{2}{3}\right )^{2}=\left( \frac{2}{3}\right )^{2} =\frac{2^2}{3^2}=\boxed{\frac{4}{9}} \\ \\ i)~~ [0,(3)]^4 =\left[\frac{3}{9}\right]^4=\left[\frac{1}{3}\right]^4 = \frac{1}{3^4}=\boxed{ \frac{1}{81}} [/tex]


[tex]\displaystyle j)~~\left(-1\frac{2}{3} \right)^{-2} = \left(-\frac{5}{3} \right)^{-2}= \left(-\frac{3}{5} \right)^{2} =\left(\frac{3}{5} \right)^{2}=\frac{3^2}{5^2} = \boxed{\frac{9}{25}} \\ \\ k)~~[-1,1(6)]^{-3}= \left[- \frac{116-11}{90}\right]^{-3}= \left[- \frac{105}{90}\right]^{-3}=\left[- \frac{90}{105}\right]^3 = \boxed{-\frac{6^3}{7^3}}[/tex]