chứng tỏ rằng
\(\dfrac{1}{2^2}\) +\(\dfrac{1}{3^2}^{^{ }}\)+...+\(\dfrac{1}{2023^2}\) < 1
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\(13\cdot28-13\cdot12+16\cdot7\)
\(=13\left(28-12\right)+16\cdot7\)
\(=13\cdot16+16\cdot7=16\left(13+7\right)=16\cdot20=320\)
Bà già đi chợ cầu Đông
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"Lợi thì có lợi, nhưng răng không còn."
A = \(\dfrac{1}{7}\) + \(\dfrac{1}{7^2}\) + \(\dfrac{1}{7^3}\) + ... + \(\dfrac{1}{7^{100}}\)
7A = 7 + \(\dfrac{1}{7}\) + \(\dfrac{1}{7^2}\) + ....+ \(\dfrac{1}{7^{100}}\)
7A - A = (7 + \(\dfrac{1}{7}\) + \(\dfrac{1}{7^2}\) +... + \(\dfrac{1}{7^{99}}\)) - (\(\dfrac{1}{7}\) + \(\dfrac{1}{7^2}\) + \(\dfrac{1}{7^3}\) + ... + \(\dfrac{1}{7^{100}}\))
6A = 7 + \(\dfrac{1}{7}\) + \(\dfrac{1}{7^2}\) + ... + \(\dfrac{1}{7^{99}}\) - \(\dfrac{1}{7}\) - \(\dfrac{1}{7^2}\) - \(\dfrac{1}{7^3}\) - ... - \(\dfrac{1}{7^{100}}\)
6A = (\(\dfrac{1}{7}\) - \(\dfrac{1}{7}\)) + (\(\dfrac{1}{7^2}\) - \(\dfrac{1}{7^2}\)) + (\(\dfrac{1}{7^3}\) - \(\dfrac{1}{7^3}\)) +...+(\(\dfrac{1}{7^{99}}\) - \(\dfrac{1}{7^{99}}\))+ (7 - \(\dfrac{1}{7^{100}}\))
6A = 0 + 0 + ... + 0 + 7 - \(\dfrac{1}{7^{100}}\)
6A = 7 - \(\dfrac{1}{7^{100}}\)
A = (7 - \(\dfrac{1}{7^{100}}\)) : 6
A = \(\dfrac{7}{6}\) - \(\dfrac{1}{6.7^{100}}\)
G = \(\dfrac{3}{5}\) + \(\dfrac{3}{5^4}\) + \(\dfrac{3}{5^7}\) + ... + \(\dfrac{3}{5^{100}}\)
53G = 75 + \(\dfrac{3}{5}\) + \(\dfrac{3}{5^4}\) +... + \(\dfrac{3}{5^{99}}\)
125G - G = (75 + \(\dfrac{3}{5}\) + \(\dfrac{3}{5^4}\) + \(\dfrac{3}{5^7}\) + ... + \(\dfrac{3}{5^{99}}\)) - (\(\dfrac{3}{5}\) + \(\dfrac{3}{5^4}\)+\(\dfrac{3}{5^7}\)+...+\(\dfrac{3}{5^{100}}\))
124G = 75 + \(\dfrac{3}{5}\) + \(\dfrac{3}{5^4}\) + \(\dfrac{3}{5^7}\)+...+ \(\dfrac{3}{5^{99}}\) - \(\dfrac{3}{5}\) - \(\dfrac{3}{5^4}\) - \(\dfrac{3}{5^7}\) - ... - \(\dfrac{3}{5^{100}}\)
124G = (75 - \(\dfrac{3}{5^{100}}\)) + (\(\dfrac{3}{5}\) - \(\dfrac{3}{5}\)) +(\(\dfrac{3}{5^4}\) - \(\dfrac{3}{5^4}\)) +...+ (\(\dfrac{3}{5^{99}}\) - \(\dfrac{3}{5^{99}}\))
124G = 75 - \(\dfrac{3}{5^{100}}\) + 0 + 0 + ... + 0
124G = 75 - \(\dfrac{3}{5^{100}}\)
G = (75 - \(\dfrac{3}{5^{100}}\)): 124
G = \(\dfrac{75}{124}\) - \(\dfrac{3}{124.5^{100}}\)
Lời giải:
\(2022A=\frac{2022^{2024}+2022}{2022^{2024}+1}=1+\frac{2021}{2022^{2024}+1}< 1+\frac{2021}{2022^{2023}+1}=\frac{2022^{2023}+2022}{2022^{2023}+1}=2022B\)
$\Rightarrow A< B$
Ta có:
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)
...
\(\dfrac{1}{2023^2}< \dfrac{1}{2022\cdot2023}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2023^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2022\cdot2023}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2023^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2023^2}< 1-\dfrac{1}{2023}< 1\)