Cho A=1/3 ×4/6 ×7/9×…..×208/210 .So sánh A với1/25 29/08/2021 Bởi Valerie Cho A=1/3 ×4/6 ×7/9×…..×208/210 .So sánh A với1/25
Đáp án: A<1/25 Giải thích các bước giải: $\begin{array}{l}Ta\,có:\frac{n}{{n + 2}} < \frac{{n – 1}}{n}\\\left( {do:{n^2} < \left( {n + 2} \right)\left( {n – 1} \right) \Rightarrow {n^2} < {n^2} + n – 2\forall n > 2} \right)\\ \Rightarrow A = \frac{1}{3}.\frac{4}{6}.\frac{7}{9}….\frac{{208}}{{210}}\\ \Rightarrow A < \frac{1}{3}.\frac{{4 – 1}}{4}.\frac{{7 – 1}}{7}….\frac{{208 – 1}}{{208}}\\ \Rightarrow A < \frac{1}{3}.\frac{3}{4}.\frac{6}{7}….\frac{{207}}{{208}}\\ \Rightarrow A.A < \frac{{1.4.7…208}}{{3.6.9…210}}.\frac{{1.3.6.9..207}}{{3.4.7.10…208}}\\ \Rightarrow {A^2} < \frac{1}{{210}}.\frac{1}{3}\\ \Rightarrow {A^2} < \frac{1}{{630}} < \frac{1}{{625}} = {\left( {\frac{1}{{25}}} \right)^2}\\ \Rightarrow A < \frac{1}{{25}}\end{array}$ Bình luận
Đáp án: A<1/25
Giải thích các bước giải:
$\begin{array}{l}
Ta\,có:\frac{n}{{n + 2}} < \frac{{n – 1}}{n}\\
\left( {do:{n^2} < \left( {n + 2} \right)\left( {n – 1} \right) \Rightarrow {n^2} < {n^2} + n – 2\forall n > 2} \right)\\
\Rightarrow A = \frac{1}{3}.\frac{4}{6}.\frac{7}{9}….\frac{{208}}{{210}}\\
\Rightarrow A < \frac{1}{3}.\frac{{4 – 1}}{4}.\frac{{7 – 1}}{7}….\frac{{208 – 1}}{{208}}\\
\Rightarrow A < \frac{1}{3}.\frac{3}{4}.\frac{6}{7}….\frac{{207}}{{208}}\\
\Rightarrow A.A < \frac{{1.4.7…208}}{{3.6.9…210}}.\frac{{1.3.6.9..207}}{{3.4.7.10…208}}\\
\Rightarrow {A^2} < \frac{1}{{210}}.\frac{1}{3}\\
\Rightarrow {A^2} < \frac{1}{{630}} < \frac{1}{{625}} = {\left( {\frac{1}{{25}}} \right)^2}\\
\Rightarrow A < \frac{1}{{25}}
\end{array}$