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a) \(A=2^{24}=\left(2^3\right)^8=8^8.\)(1)
\(B=3^{16}=\left(3^2\right)^8=9^8\)(2)
Từ (1) và (2) \(\Rightarrow A< B\)
Vậy \(A< B.\)
b) \(B=\left(0,3\right)^{30}=\left(0,3^2\right)^{15}=0,09^{15}\)(1)
\(A=\left(0,1\right)^{15}\)(2)
Từ (1) và (2) \(\Rightarrow A>B\)
Vậy \(A>B.\)
c) \(A=\left(\frac{-1}{4}\right)^8=\left(\frac{1}{4}\right)^8=\left[\left(\frac{1}{2}\right)^2\right]^8=\left(\frac{1}{2}\right)^{16}\)(1)
\(B=\left(\frac{1}{8}\right)^5=\left[\left(\frac{1}{2}\right)^3\right]^5=\left(\frac{1}{2}\right)^{15}\)(2)
Từ (1) và (2) \(\Rightarrow A>B\)
Vậy \(A>B.\)
d) \(A=102^7=102^6.102\)(1)
\(B=9^{13}=9^{12}.9=\left(9^2\right)^6.9=81^6.9\)(2)'
Từ (1) và (2) \(\Rightarrow A>B\)
Vậy \(A>B.\)
e) \(8A=8\frac{8^{18}+1}{8^{19}+1}=\frac{8^{19}+8}{8^{19}+1}=1+\frac{7}{8^{19}+1}\)(1)
\(8B=8\frac{8^{23}+1}{8^{24+1}}=\frac{8^{24}+8}{8^{24}+1}=1+\frac{7}{8^{24}+1}\)(2)
Từ (1) và (2) \(\Rightarrow8A>8B\Rightarrow A>B\)
Vậy \(A>B.\)
f) \(A=\frac{5^5}{5+5^2+5^3+5^4}=\frac{5^4}{1+5+5^2+5^3}=\frac{625}{156}>\frac{468}{156}=3.\)(1)
\(B=\frac{3^5}{3+3^2+3^3+3^4}=\frac{3^4}{1+3+3^2+3^3}=\frac{81}{40}< \frac{120}{40}=3.\)(2)
Từ (1) và (2) \(\Rightarrow A>B\)
Vậy \(A>B.\)
a, ta có A=2^24=64^4
B=3^16=81^4
Vì 64^4<81^4
Vậy 2^24<3^36
b, ta có A=0,1^15
B=0,3^30=0,09^15
Vì 0,1^15< 0,09^15
Vậy 0,1^15<0,3^30
bài 2
\(9\equiv-1\left(mod5\right)\Rightarrow9^{1945}\equiv-1^{1945}\equiv-1\left(mod5\right)\\ \)
\(2^{1930}=4^{965}\)mà \(4\equiv-1\left(mod5\right)\Rightarrow4^{965}\equiv-1^{965}\left(mod5\right)\equiv-1\left(mod5\right)\)
\(\Rightarrow9^{1945}-2^{1930}\equiv-1-\left(-1\right)\left(mod5\right)\equiv0\left(mod5\right)\Rightarrow9^{1945}-2^{1930}⋮5\)
Bài 1:
a) Ta có: \(\frac{3}{8}+\frac{-5}{6}\)
\(=\frac{3}{8}-\frac{5}{6}\)
\(=\frac{9}{24}-\frac{20}{24}\)
\(=-\frac{11}{24}\)
b) Ta có: \(\frac{15}{12}-\frac{-1}{4}\)
\(=\frac{15}{12}+\frac{1}{4}\)
\(=\frac{15}{12}+\frac{3}{12}\)
\(=\frac{18}{12}=\frac{3}{2}\)
Bài 2:
a) Ta có: \(-\frac{1}{12}-\left(2\frac{5}{8}-\frac{1}{3}\right)\)
\(=-\frac{1}{12}-\frac{21}{8}+\frac{1}{3}\)
\(=\frac{-2}{24}-\frac{63}{24}+\frac{8}{24}\)
\(=\frac{-57}{24}\)
\(=-\frac{19}{8}\)
b) Ta có: \(\frac{-5}{6}-\left(\frac{-3}{8}+\frac{1}{10}\right)\)
\(=\frac{-5}{6}+\frac{3}{8}-\frac{1}{10}\)
\(=\frac{-100}{120}+\frac{45}{120}-\frac{12}{120}\)
\(=\frac{-67}{120}\)
c) Ta có: \(-1.75-\left(\frac{-1}{9}-2\frac{1}{18}\right)\)
\(=-\frac{7}{4}+\frac{1}{9}+\frac{37}{18}\)
\(=\frac{-63}{36}+\frac{4}{36}+\frac{74}{36}\)
\(=\frac{15}{36}=\frac{5}{12}\)
ta có A= \(\frac{8^{18}+1}{8^{19} +1}\)=> 8A=\(\frac{8^{19}+8}{8^{19}+1}\)= \(\frac{\left(8^{19}+1\right)+7}{8^{19}+1}\)=\(\frac{8^{19}+1}{8^{19} +1}\)+\(\frac{7}{8^{19}+1}\) =1+\(\frac{7}{8^{19}+1}\) =\(\frac{7}{8^{19}+1}\)
B= \(\frac{8^{23}+1}{8^{24}+1}\)=> 8B=\(\frac{8^{24}+8}{8^{24}+1}\)= \(\frac{\left(8^{24}+1\right)+7}{8^{24}+1}\)=\(\frac{8^{24}+1}{8^{24}+1}\)+\(\frac{7}{8^{24}+1}\) =1+\(\frac{7}{8^{24} +1}\)=\(\frac{7}{8^{24}+1}\)
vì \(8^{19}\)<\(8^{24}\)=> \(8^{19}\)+1 >\(8^{24}\)+1 => \(\frac{7}{8^{19}+1}\)<\(\frac{7}{8^{24}+1}\)=> A<B
a) ta có \(8A=\frac{8^{19}+8}{8^{19}+1}=1+\frac{7}{8^{19}+1}\\ 8B=\frac{8^{24}+8}{8^{24}+1}=1+\frac{7}{8^{24}+1}\)
Vì \(8^{24}+1>8^{19}+1\\\frac{7}{8^{24}+1}< \frac{7}{8^{19}+1} \)
vậy 8A>8B nên A>B