Bài Toán cực khó cần giải

N

nghgh97

\[{3^{200}} = {\left( {{3^2}} \right)^{100}} = {9^{100}}\]
\[{2^{300}} = {\left( {{2^3}} \right)^{100}} = {8^{100}} < {9^{100}}\]
\[ \Rightarrow {3^{200}} > {2^{300}}\]
 
T

thong7enghiaha

$\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{48.49}+\dfrac{1}{49.50}$
$=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}$
$=\dfrac{1}{1}-\dfrac{1}{50}$
$=\dfrac{49}{50}$
 
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