Số sánh A và B: a, A=10^2015+2/10^2015-3 và B=10^2015/10^2015-5. b, A=(1/3 20/11/2021 Bởi Isabelle Số sánh A và B: a, A=10^2015+2/10^2015-3 và B=10^2015/10^2015-5. b, A=(1/32)^5 và B=(1/16)^7
Đáp án: $\begin{array}{l}a)B = \frac{{{{10}^{2015}}}}{{{{10}^{2015}} – 5}} > 1\\ \Rightarrow \frac{{{{10}^{2015}}}}{{{{10}^{2015}} – 5}} > \frac{{{{10}^{2015}} + 2}}{{{{10}^{2015}} – 5 + 2}}\\ \Rightarrow B > \frac{{{{10}^{2015}} + 2}}{{{{10}^{1015}} – 3}}\\ \Rightarrow B > A\\b)A = {\left( {\frac{1}{{32}}} \right)^5} = \frac{1}{{{{\left( {{2^5}} \right)}^5}}} = \frac{1}{{{2^{25}}}}\\B = {\left( {\frac{1}{{16}}} \right)^7} = \frac{1}{{{{\left( {{2^4}} \right)}^7}}} = \frac{1}{{{2^{28}}}}\\Vi:{2^{25}} < {2^{28}}\\ \Rightarrow \frac{1}{{{2^{25}}}} > \frac{1}{{{2^{28}}}}\\ \Rightarrow A > B\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
a)B = \frac{{{{10}^{2015}}}}{{{{10}^{2015}} – 5}} > 1\\
\Rightarrow \frac{{{{10}^{2015}}}}{{{{10}^{2015}} – 5}} > \frac{{{{10}^{2015}} + 2}}{{{{10}^{2015}} – 5 + 2}}\\
\Rightarrow B > \frac{{{{10}^{2015}} + 2}}{{{{10}^{1015}} – 3}}\\
\Rightarrow B > A\\
b)A = {\left( {\frac{1}{{32}}} \right)^5} = \frac{1}{{{{\left( {{2^5}} \right)}^5}}} = \frac{1}{{{2^{25}}}}\\
B = {\left( {\frac{1}{{16}}} \right)^7} = \frac{1}{{{{\left( {{2^4}} \right)}^7}}} = \frac{1}{{{2^{28}}}}\\
Vi:{2^{25}} < {2^{28}}\\
\Rightarrow \frac{1}{{{2^{25}}}} > \frac{1}{{{2^{28}}}}\\
\Rightarrow A > B
\end{array}$