9^12va27^7 So sanh nha
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\(A< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{5}{5}=1=B\)
a/
\(\frac{2001}{2004}=\frac{2004-3}{2004}=1-\frac{3}{2004}=1-\frac{1}{668}.\)
\(\frac{39}{40}=\frac{40-1}{40}=1-\frac{1}{40}\)
Ta có \(40< 668\Rightarrow\frac{1}{40}>\frac{1}{668}\Rightarrow1-\frac{1}{40}< 1-\frac{1}{668}\Rightarrow\frac{39}{40}< \frac{2001}{2004}\)
b/
\(A< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=1=B\)
\(\frac{-4}{3}=\frac{-28}{21};\frac{9}{-7}=\frac{-27}{21}\Rightarrow\frac{-4}{3}
ta có \(\frac{3}{7}=\frac{27}{63}\); \(\frac{4}{9}=\frac{28}{63}\)
vì \(\frac{27}{63}< \frac{28}{63}\)
nên\(\frac{3}{7}< \frac{4}{9}\)
\(8^5=\left(2^3\right)^5=2^{15}=2\times2^{14}\)
\(3\times4^7=3\times\left(2^2\right)^7=3\times2^{14}\)
vì 3>2 nên \(3\times2^{14}>2\times2^{14}\)hay \(3\times4^7>8^5\)
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
=> \(\frac{1}{2^{35}}>\frac{1}{2^{36}}\)
=> \(\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)
\(\left(\frac{1}{32}\right)^7=\left[\left(\frac{1}{2}\right)^5\right]^7=\left(\frac{1}{2}\right)^{35}\)và \(\left(\frac{1}{16}\right)^9=\left[\left(\frac{1}{2}\right)^4\right]^9=\left(\frac{1}{2}\right)^{32}\)
Mà:\(\left(\frac{1}{2}\right)^{35}>\left(\frac{1}{2}\right)^{32}\Rightarrow\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)
Ta có : \(A-1=\frac{9^{11}+1}{9^{11}-7}-1=\frac{8}{9^{11}-7}\) ; \(B-1=\frac{9^{12}+3}{9^{12}-5}-1=\frac{8}{9^{12}-5}\)
Cần so sánh : \(9^{11}-7\) và \(9^{12}-5\)
Ta viết : \(9^{12}-5=9^{11}.9-5=9^{11}.\left(1+8\right)-5=\left(9^{11}-7\right)+\left(8.9^{11}+2\right)\)
Xét : \(\left(9^{12}-5\right)-\left(9^{11}-7\right)=\left(9^{11}-7\right)+\left(8.9^{11}+2\right)-\left(9^{11}-7\right)=8.9^{11}+2>0\)
\(\Rightarrow9^{12}-5>9^{11}-7\)
Do đó : \(B-1>A-1\Rightarrow B< A\)
9^12=(3^2)^12=3^24
27^7=(3^3)^7=3^21
vi 3^24>3^27 nen 9^12>27^7
chuc em hoc gioi
Ta có : \(9^{12}=\left(3^2\right)^{12}=3^{2.12}=3^{24}\)
\(27^7=\left(3^3\right)^7=3^{3.7}=3^{21}\)
Vì \(24>21\)\(=>3^{21}< 3^{24}=>27^7< 9^{12}\)
Vậy \(9^{12}>27^7\)