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\(\frac{7}{10}< \frac{6}{7}< \frac{48}{55}< \frac{12}{11}< \frac{8}{7}< \frac{7}{5}< \frac{3}{2}< \frac{9}{4}\)
Bài làm
a ) \(A=\frac{9^{99}+1}{9^{100}+1}=\frac{9^{100}+1}{9^{100}+1}-\frac{9}{9^{100}+1}\)
= \(1-\frac{9}{9^{100}+1}\)
\(B=\frac{10^{98}-1}{10^{99}-1}=\frac{10^{99}-1}{10^{99}-1}-\frac{10}{10^{99}-1}\)
= \(1-\frac{10}{10^{99}-1}\)
Vì \(\frac{9}{9^{100}+1}>\frac{10}{10^{99}-1}\)
nên \(1-\frac{9}{9^{100}+1}< 1-\frac{10}{10^{99}-1}\)
\(\Rightarrow A< B\)
Bài làm
b ) \(A=\frac{5^{10}}{1+5+5^2+.....+5^9}=\frac{1+5+5^2+.....+5^9}{1+5+5^2+.....+5^9}+\frac{1+5+5^2+.....+5^8-5^9.4}{1+5+5^2+.....+5^9}\)
= \(1+\frac{1+5+5^2+.....+5^8+5^9.4}{1+5+5^2+.....+5^9}=1+5^9.3\)
\(B=\frac{6^{10}}{1+6+6^2+.....+6^9}=\frac{1+6+6^2+.....+6^9}{1+6+6^2+.....+6^9}+\frac{1+6+6^2+.....+6^8+6^9.5}{1+6+6^2+.....+6^9}\)
= \(1+\frac{1+6+6^2+.....+6^8+6^9.5}{1+6+6^2+.....+6^9}=1+6^9.4\)
Vì \(1+5^9.3< 1+6^9.4\)
nên A < B
\(\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{3^{10}.\left(-5\right)^{20}.\left(-5\right)}{\left(-5\right)^{20}.3^{10}.3^2}=\frac{-5}{3^2}=-\frac{5}{9}\)
\(B=\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^9.33+3^9.15}{3^9.2^4}\)
\(=\frac{3^9\left(33+15\right)}{3^9.2^4}=\frac{3^9.48}{3^9.16}\)
\(=\frac{48}{16}=3\)
\(B=\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
\(=\frac{3^{10}.\left(11+5\right)}{3^9.8}\)
\(=\frac{3^{10}.16}{3^9.8}\)
\(=\frac{3.2}{1}\)
\(=6\)
\(\frac{5^{12}.3^9-5^{10}.3^{11}}{5^{10}.3^{10}}\)\(=\frac{5^{10}.3^9(5^2-3^2)}{5^{10}.3^{10}}\)\(=\frac{5^2-3^2}{3}\)\(=\frac{16}{3}\)
Chúc bạn học tốt!