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1: \(=\dfrac{3}{4}+\dfrac{5}{4}\cdot\dfrac{8}{3}-\dfrac{1}{4}\cdot\dfrac{5}{6}=\dfrac{3}{4}+\dfrac{10}{3}-\dfrac{5}{24}\)
\(=\dfrac{18}{24}+\dfrac{80}{24}-\dfrac{5}{24}=\dfrac{93}{24}=\dfrac{31}{8}\)
2: \(=\left(7+\dfrac{23}{27}-\dfrac{23}{27}\right)+\left(\dfrac{11}{25}+\dfrac{14}{25}\right)+3.25\)
\(=7+1+3.25=8+3.25=11.25\)
3: \(=\left(\dfrac{1}{9}\cdot9\right)^{2005}-4^2=1-16=-15\)
4: \(=2\cdot\dfrac{9}{4}-\dfrac{7}{2}=\dfrac{9}{2}-\dfrac{7}{2}=1\)
5: \(=\dfrac{15}{2}\cdot\dfrac{-3}{5}+\dfrac{5}{2}\cdot\dfrac{-3}{5}=\dfrac{-3}{5}\cdot\left(\dfrac{15}{2}+\dfrac{5}{2}\right)=\dfrac{-3}{5}\cdot10=-6\)
6: \(=\left(\dfrac{6}{10}+\dfrac{5}{10}\right)^2=\left(\dfrac{11}{10}\right)^2=\dfrac{121}{100}\)
7: \(=\dfrac{1}{2}\cdot\dfrac{-7}{2}=\dfrac{-7}{4}\)
\(8A=\frac{8^{19}+8}{8^{19}+1}=1+\frac{7}{8^{19}+1}\)
\(8B=\frac{8^{24}+8}{8^{24}+1}=1+\frac{7}{8^{24}+1}\)
\(\text{Vì }\frac{7}{8^{19}+1}>\frac{7}{8^{24}+1}\)
\(\Rightarrow8A>8B\)
\(\Rightarrow A>B\)
\(\text{Câu B làm tương tự nhé}\)
bài 2
làm câu B;C nha
B)
\(27^3=\left(3^3\right)^3=3^9\)
\(9^5=\left(3^2\right)^5=3^{10}\)
vì \(10>9\)
\(=>9^5>27^3\)
C)
\(\left(\frac{1}{8}\right)^6=\left(\frac{1}{2^3}\right)^6=\frac{1^6}{2^{18}}=\frac{1}{2^{18}}\)
\(\left(\frac{1}{32}\right)^4=\left(\frac{1}{2^5}\right)^4=\frac{1^4}{2^{20}}=\frac{1}{2^{20}}\)
vì \(2^{18}< 2^{20}\)
\(=>\frac{1}{2^{18}}>\frac{1}{2^{20}}\)
\(=>\left(\frac{1}{8}\right)^6>\left(\frac{1}{32}\right)^4\)
\(\text{A.}\frac{32^3.9^5}{8^3.6^6}=\frac{\left(2^5\right)^3.\left(3^2\right)^5}{\left(2^3\right)^3.\left(2.3\right)^6}=\frac{2^{15}.3^{10}}{2^9.2^6.3^6}=\frac{3^{10}}{3^6}=3^4=81\)
\(\text{B.}\frac{\left(5^5-5^4\right)^3}{50^6}=\frac{2500^3}{50^6}=\frac{\left(50^2\right)^3}{50^6}=\frac{50^6}{50^6}=1\)
Bài 2:
\(\text{A.Ta có:}\)
\(5^6=\left(5^3\right)^2=125^2\)
\(\left(-2\right)^{14}=2^{14}=\left(2^7\right)^2=128^2\)
Vì \(125< 128\)
\(\Rightarrow125^2< 128^2\)
\(\Rightarrow5^6< \left(-2\right)^{14}\)
\(\text{B.Ta có:}\)
\(9^5=\left(3^2\right)^5=3^{10}\)
\(27^3=\left(3^3\right)^3=3^9\)
Vì \(9< 10\)
\(\Rightarrow3^9< 3^{10}\)
\(\Rightarrow27^3< 9^5\)
\(\text{C.Ta có:}\)
\(\left(\frac{1}{8}\right)^6=\left[\left(\frac{1}{2}\right)^3\right]^6=\left(\frac{1}{2}\right)^{18}\)
\(\left(\frac{1}{32}\right)^4=\left[\left(\frac{1}{2}\right)^5\right]^4=\left(\frac{1}{2}\right)^{20}\)
Vì \(18< 20\)
\(\Rightarrow\left(\frac{1}{2}\right)^{18}< \left(\frac{1}{2}\right)^{20}\)
\(\Rightarrow\left(\frac{1}{8}\right)^6< \left(\frac{1}{32}\right)^4\)
A= \(\frac{25^3.5^3}{6.5^{10}}\)= \(\frac{\left(5^2\right)^3.5^3}{6.5^{10}}\)= \(\frac{5^6.5^3}{6.5^{10}}\)= \(\frac{5^9}{6.5^{10}}\)= \(\frac{5}{6}\)
B = \(\frac{2^5.6^3}{8^2.9^2}\)= \(\frac{2^5.\left(2.3\right)^3}{\left(2^3\right)^2.\left(3^2\right)^2}\)=\(\frac{2^5.2^3.3^3}{2^6.3^4}\)= \(\frac{2^8.3^3}{2^6.3^4}\)= \(\frac{2^2}{3}\)= \(\frac{4}{3}\)
C = \(\frac{15^3+5.15^2-5^3}{18^3+6.18^2-6^3}\)= \(\frac{5^3.3^3+5.5^2.3^2-5^3}{6^3.3^3+6.6^2.3^2-6^3}\)= \(\frac{5^3+5^3.3^2-5^3}{6^3.3^3+6^3.3^2-6^3}\)= \(\frac{5^3.\left(1+3^2-1\right)}{6^3.\left(3^3+3^2-1\right)}\)= \(\frac{5^3.9}{6^3.35}\)
=\(\frac{5^3.3^2}{2^3.3^3.7.5}\)
= \(\frac{25}{168}\)
D = \(\frac{\left(7^4-7^3\right)^2}{49^3}\)= \(\frac{[7^3\left(7-1\right)]^2}{\left(7^2\right)^3}\)= \(\frac{7^6.6^2}{7^6}\)= \(36\)