Ta có : 

\(\left(\frac{1}{32}\right)^7=\frac{1^7}{32^7}=\frac{1}{\left(2^5\right)^7}=\frac{1}{2^{5.7}}=\frac{1}{2^{35}}\)

\(\left(\frac{1}{16}\right)^9=\frac{1^9}{16^9}=\frac{1}{\left(2^4\right)^9}=\frac{1}{2^{4.9}}=\frac{1}{2^{36}}\)

Vì \(\frac{1}{2^{35}}>\frac{1}{2^{36}}\) ( cùng tử, mẫu nào bé hơn thì phân số đó lớn hơn ) nên \(\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)

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

Chúc bạn học tốt ~ 

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

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

DO :  \(\frac{1}{2^{35}}>\frac{1}{2^{36}}\)\(\Rightarrow\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)

Tk mk nha !!! 

> nha bn

b)Có \(63^7< 64^7\)

\(64^7=\left(2^6\right)^7=2^{42}\)

\(16^{12}=\left(2^4\right)^{12}=2^{48}\)

Mà \(2^{42}< 2^{48}\Rightarrow63^7< 64^7< 16^{12}\Rightarrow63^7< 16^{12}\)

bai toan nay kho

\(\frac{2011.4023+2012}{2012.4023-2011}=\frac{2011.4023+2011+1}{2012.4023-2012-1}=\frac{2011.4023+2011.1+1}{2012.4023-2012.1-1}\)

\(=>\frac{2012.4023+2012.1+1}{2012.4023-2012.1-1}=\frac{2012.\left(4023+1\right)+1}{2012.\left(4023-1\right)-1}\)

\(=\frac{4023+1+1}{4023-1-1}=\frac{4023+2}{4023-2}=\frac{4025}{4021}\)

Vì 4025 > 4021 ( tử số lớn hơn mẫu số ) nên suy ra : 4025/4021 >1

<br class="Apple-interchange-newline"><div id="inner-editor"></div>=>2012.4023+2012.1+12012.4023−2012.1−1 =2012.(4023+1)+12012.(4023−1)−1 

=4023+1+14023−1−1 =4023+24023−2 =40254021 

Vì 4025 > 4021 ( tử số lớn hơn mẫu số ) nên suy ra : 4025/4021 >1

Ta có: (+)  (1/32)^7   =  [(1/2)^5]^7   =(1/2)^35

          (+)  (1/16)^9=    [(1/2)^4]^9   =(1/2)^36

          Vì 35 <36

   => (1/2)^35 > (1/2)^36

  => (1/32)^7 > (1/16)^9

Bài Toán khó đây

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

duyệt đi

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

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

Vì: 2³⁵ < 2³⁶ nên \(\frac{1}{2^{35}}>\frac{1}{2^{36}}\)

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

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

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

Vì \(2^{35}\frac{1}{2^{36}}\)

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

(1/32)^7 = (1/2^5)^7 =(1/2)^35 > ( 1/2 ) ^36 = (1/2^4 )^9 = ( 1/ 16 ) ^9