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a) \(63^7\)và \(16^{12}\)
Có \(63^7< 64^7=\left(2^6\right)^7=2^{42}\)
\(16^{12}=\left(2^4\right)^{12}=2^{48}\)
Mà \(2^{42}< 2^{48}\Rightarrow63^7< 64^7< 16^{12}\)=) \(63^7< 16^{12}\)
b) \(17^{14}\)và \(31^{11}\)
Có \(17^{14}>16^{14}=\left(2^4\right)^{14}=2^{56}\)
\(31^{11}< 32^{11}=\left(2^5\right)^{11}=2^{55}\)
Vì \(2^{56}>2^{55}\Rightarrow17^{14}>16^{14}>32^{11}>31^{11}\)
=) \(17^{14}>31^{11}\)
c) \(2^{67}\)và \(5^{21}\)
Có \(5^{21}< 8^{21}=\left(2^3\right)^{21}=2^{63}\)
Vì \(2^{67}>2^{63}\Rightarrow2^{67}>8^{21}>5^{21}\)
=) \(2^{67}>5^{21}\)
Bài 1:
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + (\(\frac45\) - \(\frac{3}{17}\) + \(\frac13\)) - \(\frac17\) + (- \(\frac{14}{30}\))
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + \(\frac45\) - \(\frac{3}{17}\) + \(\frac13\) - \(\frac17\) - \(\frac{14}{30}\)
A = (\(\frac15\) + \(\frac45\)) + (\(\frac{3}{17}\) - \(\frac{3}{17}\)) - (\(\frac43-\frac13\)) - \(\frac{30}{210}\) - \(\frac{98}{210}\)
A = 1 + 0 - 1 - (\(\frac{30}{210}+\frac{98}{210}\))
A = 1 - 1 - \(\frac{228}{210}\)
A = 0 - \(\frac{128}{210}\)
A = - \(\frac{64}{105}\)
Bài 2:
B= (\(\frac58\) - \(\frac{4}{12}\) + \(\frac32\)) - (\(\frac58\) + \(\frac{9}{13}\)) - (\(\frac{-3}{2}\)) + \(\frac{7}{-15}\)
B = \(\frac58\) - \(\frac{4}{12}\) + \(\frac32\) - \(\frac58\) - \(\frac{9}{13}\) + \(\frac32\) - \(\frac{7}{15}\)
B = (\(\frac58\) - \(\frac58\)) + (\(\frac32\) + \(\frac32\)) - (\(\frac13\) + \(\frac{9}{13}\) + \(\frac{7}{15}\))
B = 0 + 3 - (\(\frac{65}{195}\) + \(\frac{135}{195}\) + \(\frac{91}{195}\))
B = 3 - (\(\frac{200}{195}\) + \(\frac{91}{195}\))
B = 3 - \(\frac{97}{65}\)
B = \(\frac{195}{65}\) - \(\frac{97}{65}\)
B = \(\frac{98}{65}\)
a/ \(63^7< 64^7=\left(4^3\right)^7=4^{21}\)
\(16^{12}=\left(4^2\right)^{12}=4^{24}\)
Suy ra \(63^7< 4^{21}< 4^{24}=16^{12}\)
Vậy \(63^7< 16^{12}\)
b) \(\frac{\frac{2}{3}+\frac{5}{7}+\frac{4}{21}}{\frac{5}{6}+\frac{11}{7}-\frac{7}{21}}\)
\(=\frac{\frac{29}{21}+\frac{4}{21}}{\frac{101}{42}-\frac{7}{21}}\)
\(=\frac{\frac{11}{7}}{\frac{29}{14}}\)
\(=\frac{22}{29}.\)
Chúc bạn học tốt!
a: \(A=\dfrac{-13}{21}=\dfrac{-26}{42}\)
\(B=\dfrac{-9}{14}=\dfrac{-27}{42}\)
mà -26>-27
nên A>B
b: \(A=\dfrac{99}{101}=1-\dfrac{2}{101}\)
\(B=\dfrac{2011}{2013}=1-\dfrac{2}{2013}\)
mà 2/101>2/2013
nên A<B
a, <
b, <
c,<
d, =
e,>
g, <
Bn bấm mấy tính r so các số ở cuối, giả sử = 5...,... x\(10^4\) còn số còn lại là 5...,...x\(10^5\) thì số kia to hơn do x\(10^5\) > x\(10^4\) ( mẹo thôi chứ m ko bt cách làm dạng bài này)
a: \(5^{28}=\left(5^2\right)^{14}=25^{14}\)
mà 25<26
nên \(5^{28}=25^{14}<26^{14}\)
c: \(31^{11}<32^{11}=\left(2^5\right)^{11}=2^{55}\)
\(17^{14}>16^{14}=\left(2^4\right)^{14}=2^{56}\)
mà \(2^{56}>2^{55}\)
nên \(17^{14}>31^{11}\)
d: \(64^7=\left(4^3\right)^7=4^{3\cdot7}=4^{21}\)
e: \(2^{91}=\left(2^{13}\right)^7=8192^7\)
\(5^{35}=\left(5^5\right)^7=3125^7\)
mà 8192>3125
nên \(2^{91}>5^{35}\)
g: \(21^{12}=\left(21^3\right)^4=9261^4>54^4\)