Question

1) so sanh

(1/32)^7 va (1/16)^9

Answer 1

Ta có:

\(\left(\frac{1}{32}\right)^7=\frac{1^7}{32^7}=\frac{1}{\left(2^5\right)^7}=\frac{1}{2^{35}}\)

\(\left(\frac{1}{16}\right)^9=\frac{1^9}{16^9}=\frac{1}{\left(2^4\right)^9}=\frac{1}{2^{36}}\)

Vì 2³⁵ < 2³⁶

=> \(\frac{1}{2^{35}}>\frac{1}{2^{36}}\)

=> \(\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)

Answer 2

\(\left(\frac{1}{32}\right)^7=\left[\left(\frac{1}{2}\right)^5\right]^7=\left(\frac{1}{2}\right)^{35}\)và \(\left(\frac{1}{16}\right)^9=\left[\left(\frac{1}{2}\right)^4\right]^9=\left(\frac{1}{2}\right)^{32}\)

Mà:\(\left(\frac{1}{2}\right)^{35}>\left(\frac{1}{2}\right)^{32}\Rightarrow\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)

Answer 3

\(\left(\frac{1}{32}\right)^7=\left(\frac{1}{2}\right)^{35}\)

\(\left(\frac{1}{16}\right)^9=\left(\frac{1}{2}\right)^{32}\)

Do \(\left(\frac{1}{2}\right)^{35}>\left(\frac{1}{2}\right)^{32}\Rightarrow\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)