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a)\(4^{72}=\left(4^3\right)^{24}=64^{24}\)
\(8^{48}=\left(8^2\right)^{24}=64^{24}\)
\(\Rightarrow4^{72}=8^{48}\)
a) \(4^{72}=\left(2^2\right)^{72}=2^{144}\)
\(8^{48}=\left(2^3\right)^{48}=2^{144}\)
mà \(2^{144}=2^{144}\)=> \(4^{72}=8^{48}\)
b) \(2^{252}=\left(2^2\right)^{126}=4^{126}\)
mà \(4^{126}< 5^{127}\)=> \(5^{127}>2^{252}\)
Câu 1 :
Ta có : \(A=\frac{10^{100}+1}{10^{101}+1}\)
\(\Rightarrow10A=\frac{10^{101}+10}{10^{101}+1}=\frac{10^{101}+1+9}{10^{101}+1}=1+\frac{9}{10^{101}+1}\)
Ta có : \(B=\frac{10^{101}+1}{10^{102}+1}\)
\(10B=\frac{10^{102}+10}{10^{102}+1}=\frac{10^{102}+1+9}{10^{102}+1}=1+\frac{9}{10^{102}+1}\)
Vì 10101+1<10102+1
\(\Rightarrow\frac{9}{10^{101}+1}>\frac{9}{10^{102}+1}\)
\(\Rightarrow1+\frac{9}{10^{101}+1}>1+\frac{9}{10^{102}+1}\)
\(\Rightarrow\)10A>10B
\(\Rightarrow\)A>B
Vậy A>B.
Câu 2 :
Ta có : \(E=\frac{2000+2001}{2001+2002}=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)
Vì 2001<2001+2002 và 2002<2001+2002
\(\Rightarrow\hept{\begin{cases}\frac{2000}{2001}>\frac{2000}{2001+2002}\\\frac{2001}{2002}>\frac{2001}{2001+2002}\end{cases}}\)
\(\Rightarrow C>E\)
Vậy C>E.
Câu 1,
\(S=1+2+2^2+...+2^7\)
\(=\left(1+2\right)+2^2\left(1+2\right)+2^4\left(1+2\right)+2^6\left(1+2\right)\)
\(=3+2^2.3+2^4.3+2^6.3\)
\(=3\left(1+2^2+2^4+2^6\right)⋮3\)
Nên S chia hết cho 3
Câu 2 ,
\(A=5+5^2+5^3+...+5^{20}\)
\(=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{19}\left(1+5\right)\)
\(=5.6+5^3.6+...+5^{19}.6\)
\(=6\left(5+5^3+...+5^{19}\right)⋮6\)
Nên A chia hết cho 6
Ta có :
\(A=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)
\(A=5\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)
\(A=5\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(A=5\left(1-\frac{1}{31}\right)\)
\(A=5.\frac{30}{31}\)
\(A=\frac{150}{31}>1\)
\(\Rightarrow\)\(A>1\)
Vậy \(A>1\)
Chúc bạn học tốt ~
Ko cần dài dòng vậy đâu,A=\(\frac{5^2}{1.6}+\left(\frac{5^2}{6.11}+\frac{5^2}{11.16}+...+\frac{5^2}{26.31}\right)\)
Ta thấy \(\frac{5^2}{1.6}>1\)và tổng trong ngoặc >0 nên =>A>1
(\(\frac{12}{19}.\frac{19}{12}\))(\(\frac{-13}{17}.\frac{17}{13}\))\(\frac{7}{15}\)
= 1 . (-1) . \(\frac{7}{15}\)
= \(-\frac{7}{15}\)
Ta có: \( \left(\frac{12}{19}\times\frac{19}{12}\right)\times\left(\frac{-13}{17}\times\frac{17}{13}\right).\frac{19}{12}\)
=1.(-1).\(\frac{19}{12}\)
=(-1).\(\frac{19}{12}\)
=\(-\frac{19}{12}\)
a) Ta có \(10^{2m}+18=10^{2m}-10^m+10^m+18\)
\(=10^m.\left(10^m-1\right)+\left(10^m-1\right)+19⋮19\)