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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
A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(=\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)
\(=\frac{3}{5}+\frac{2}{5}=1\)
b) B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{1}{3.2}-\frac{5.2}{7.3}\)
\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)
\(=\frac{7}{42}-\frac{20}{42}\)
\(=-\frac{13}{42}\)
Bài 1:
a) \(\left(\frac{9}{25}-2.18\right):\left(3\frac{4}{5}+0,2\right)\)
\(=\left(\frac{9}{25}-36\right):\left(\frac{19}{5}+\frac{1}{5}\right)\)
\(=\left(\frac{9}{25}-\frac{900}{25}\right):4\)
\(=-\frac{891}{25}.\frac{1}{4}\)
\(=-\frac{891}{100}\)
b) \(\frac{3}{8}.19\frac{1}{3}-\frac{3}{8}.33\frac{1}{3}\)
\(=\frac{3}{8}.\frac{58}{3}-\frac{3}{8}.\frac{100}{3}\)
\(=\frac{3}{8}\left(\frac{58}{3}-\frac{100}{3}\right)\)
\(=\frac{3}{8}\left(-\frac{42}{3}\right)\)
\(=\frac{3}{8}.\left(-14\right)\)
\(=-\frac{21}{4}\)
c) \(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=\frac{27}{23}+\frac{5}{21}-\frac{4}{23}+\frac{1}{2}+\frac{16}{21}\)
\(=\frac{27}{23}+\frac{5}{21}+\left(-\frac{4}{23}\right)+\frac{1}{2}+\frac{16}{21}\)
\(=\left[\frac{27}{23}+\left(-\frac{4}{23}\right)\right]+\left(\frac{5}{21}+\frac{16}{21}\right)+\frac{1}{2}\)
\(=1+1=2\)
d) \(\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{4}{5}\)
\(=\frac{21}{47}+\frac{9}{45}+\frac{26}{47}+\frac{36}{45}\)
\(=\left(\frac{21}{47}+\frac{26}{47}\right)+\left(\frac{9}{45}+\frac{36}{45}\right)\)
\(=1+1=2\)
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: \(-1.75-\left(\frac{-1}{9}-2\frac{1}{18}\right)\)
\(=-\frac{7}{4}+\frac{1}{9}+\frac{37}{18}\)
\(=\frac{-63}{36}+\frac{4}{36}+\frac{74}{36}\)
\(=\frac{15}{36}=\frac{5}{12}\)
Bài 1:
a) Ta có: \(25\cdot\left(\frac{-1}{5}\right)^3+\frac{1}{5}-2\cdot\left(\frac{-1}{2}\right)^2-\frac{1}{2}\)
\(=25\cdot\frac{-1}{125}+\frac{1}{5}-2\cdot\frac{1}{4}-\frac{1}{2}\)
\(=-\frac{1}{5}+\frac{1}{5}-\frac{1}{2}-\frac{1}{2}\)
\(=\frac{-2}{2}=-1\)
b) Ta có: \(35\frac{1}{6}:\left(\frac{-4}{5}\right)-46\frac{1}{6}:\left(\frac{-4}{5}\right)\)
\(=\frac{211}{6}\cdot\frac{-5}{4}-\frac{277}{6}\cdot\frac{-5}{4}\)
\(=\frac{-5}{4}\cdot\left(\frac{211}{6}-\frac{277}{6}\right)\)
\(=\frac{-5}{4}\cdot\left(-11\right)=\frac{55}{4}\)
c) Ta có: \(\left(\frac{-3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)
\(=\frac{-7}{20}\cdot\frac{7}{3}+\frac{7}{20}\cdot\frac{7}{3}\)
\(=\frac{7}{3}\cdot\left(-\frac{7}{20}+\frac{7}{20}\right)=\frac{7}{3}\cdot0=0\)
d) Ta có: \(\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{18}\right)+\frac{7}{8}\cdot\left(\frac{1}{36}-\frac{5}{12}\right)\)
\(=\frac{7}{8}\cdot6+\frac{7}{8}\cdot\frac{-7}{18}\)
\(=\frac{7}{8}\cdot\left(6+\frac{-7}{18}\right)\)
\(=\frac{7}{8}\cdot\frac{101}{18}=\frac{707}{144}\)
e) Ta có: \(\frac{1}{6}+\frac{5}{6}\cdot\frac{3}{2}-\frac{3}{2}+1\)
\(=\frac{1}{6}+\frac{15}{12}-\frac{3}{2}+1\)
\(=\frac{2}{12}+\frac{15}{12}-\frac{18}{12}+\frac{12}{12}\)
\(=\frac{11}{12}\)
f) Ta có: \(\left(-0,75-\frac{1}{4}\right):\left(-5\right)+\frac{1}{15}-\left(-\frac{1}{5}\right):\left(-3\right)\)
\(=\left(-1\right):\left(-5\right)+\frac{1}{15}-\frac{1}{15}\)
\(=\frac{1}{5}\)
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