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\(\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)
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
(1\32)^7 =(1\2)^35 >(1\2)^36=(1\16)^9
=>(1\32)^7 > (1\16)^9
Vậy (1\32)^7 > (1\16)^9
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 !!!
Ta có: A = \(\frac{-2}{11}+\frac{6}{7}+\frac{1}{2}+\frac{-9}{11}+\frac{1}{7}\)
A = \(\left(\frac{-2}{11}+\frac{-9}{11}\right)+\left(\frac{6}{7}+\frac{1}{7}\right)+\frac{1}{2}\)
A = \(-1+1+\frac{1}{2}\)
A = \(\frac{1}{2}\)
B = \(\left(\frac{9}{16}+\frac{8}{27}\right)+\left(1+\frac{7}{16}+\frac{-19}{27}\right)\)
B = \(\frac{9}{16}+\frac{8}{27}+1+\frac{7}{16}-\frac{19}{27}\)
B = \(\left(\frac{9}{16}+\frac{7}{16}\right)+1+\left(\frac{8}{27}-\frac{19}{27}\right)\)
B = \(1+1-\frac{11}{27}\)
B = \(\frac{43}{27}\)
Mà 1/2 < 43/27 (Vì 1/2 < 1; 43/27 > 1)
=> A < B
Giải
\(A=\frac{-2}{11}+\frac{6}{7}+\frac{1}{2}+\frac{-9}{11}+\frac{1}{7}\)
\(\Leftrightarrow A=\left(\frac{-2}{11}+\frac{-9}{11}\right)+\left(\frac{6}{7}+\frac{1}{7}\right)+\frac{1}{2}\)
\(\Leftrightarrow A=\frac{-11}{11}+\frac{7}{7}+\frac{1}{2}\)
\(\Leftrightarrow A=-1+1+\frac{1}{2}\)
\(\Leftrightarrow A=\frac{1}{2}< 1\left(1\right)\)
\(B=\left(\frac{9}{16}+\frac{8}{27}\right)+\left(1+\frac{7}{16}+\frac{-19}{27}\right)\)
\(\Leftrightarrow B=\left(\frac{9}{16}+\frac{7}{16}\right)+\left(\frac{8}{27}+\frac{-19}{27}\right)+1\)
\(\Leftrightarrow B=\frac{16}{16}+\frac{-11}{27}+1\)
\(\Leftrightarrow B=1+\frac{-11}{27}+1\)
\(\Leftrightarrow B=2+\frac{-11}{27}\)
\(\Leftrightarrow B=\frac{43}{27}\)\(>1\left(2\right)\)
Từ (1) và (2) suy ra A < B