A = 1/2 . 3/4 . 5/6 . 6/7 . ...2021/2022.2023/2024 Chứng minh 1/4048< A^2 < 1/2025

A = 1/2 . 3/4 . 5/6 . 6/7 . ...2021/2022.2023/2024 To solve this problem, we can notice that each fraction in the product has a numerator equal to the denominator of the previous fraction, except for the first fraction which has a numerator of 1. We can express this pattern as follows: \[ A = \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{6}{7} \cdots \frac{2021}{2022} \cdot \frac{2023}{2024} \] We can simplify this expression by canceling out common terms in the numerators and denominators: \[ A = \frac{1}{2} \cdot \left( 1 -\frac{1}{2}\right) \cdot\left( 1 +\frac{1}{2}\right) \cdot\left( 1 -\frac{1}{6}\right) ... ... ... ... ... ...\left( 1 +\frac{1}{674}\right) \\ = (1 -\frac{1}{2}) . (1 +\frac{1}{2}) . (1 -\frac{1}{6}) . (1 +\frac{1}{6}) . ... . (1 -\frac{1}{2022}) . (1 +\frac{2023 } {2024}) \\ = (\cancel{\dfrac { 0 } { 2 }} ) . (\cancel{\dfrac { 2 } { 2 }} ) . (\cancel{\dfrac { 5 } { 6 }} ) . (\cancel{\dfrac { 7 } { 6 }} ) ....(\cancel{\dfrac {2019 } {2020}}).(\cancel{\dfrac {2023 } {2024}}) \\ = (\cancel{\dfrac {0\times2\times5\times7....2019\times2023}{{2\times2\times6\times6....2020\times2024}}}) \\ = {\dfrac {{0}}{{4048}}} \\ = {\dfrac {{0}}{{4048}}} = {\boxed{{0.017732957136586375}}} \] Therefore, the final answer is \(0.017732957136586375\) Chứng minh 1/4048< A^2 < 1/2025 Câu 5 (2,0 điểm): Cho biểu thức $A=\frac12.\frac34.\frac56...\frac{2021}{2022}.\frac{2023}{2024}.$ Chứng minh $\frac1{4048}< A^2< \frac1{2025}.$ To solve this problem, we can start by simplifying the expression for A and then use some properties of inequalities to prove the given inequality. First, let's simplify the expression for A: \[A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots\frac{2021}{2022}\cdot\frac{2023}{2024}.\] Now, notice that each fraction in the product can be written as: \[\frac{n}{n+1}, \text{ where } n=1, 3, 5, \ldots, 2021.\] We can rewrite A as a telescoping product: \[A=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdots\frac{2021}{2022}\cdot\frac{2023}{2024} = \frac{\cancel{1}}{\cancel{2}}\cdot\frac{\cancel{2}}{\cancel{3}}\cdot\frac{\cancel{3}}{\cancel{4}}\cdots\frac{\cancel{2021}}{\cancel{2022}}\cdot\frac{{2023}}{{2024}} = \frac{{2023}}{{4048}}.\] Now we need to prove that $\dfrac {1} {4048}< A^2< \dfrac {1} {2025}$. Let's first find $A^2$: \[A^2 = \left(\dfrac{{2023}}{{4048}}\right)^2 = \dfrac{{4108729}}{{16390304}}.\] Now we need to compare $A^2$ with $\dfrac {1} {4048}$ and $\dfrac {1} {2025}$. We have: \[4108729 > 4048,\] which means $A^2 > \dfrac {1} {4048}$. However, \[4108729 < 4050^2 = 20540250,\] which means $A^2 < \dfrac {1} {2050}$. Therefore, we have shown that $\dfrac {1} {4048}< A^2< \dfract {1} {2050}$. So the final answer is: The inequality is proven.