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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
1 tháng 12 2023
A = \(\dfrac{\dfrac{2022}{1}+\dfrac{2021}{2}+\dfrac{2020}{3}+...+\dfrac{1}{2022}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}}\)
Xét TS = \(\dfrac{2022}{1}\) + \(\dfrac{2021}{2}\) \(\dfrac{2020}{3}\) +... + \(\dfrac{1}{2022}\)
TS = (1 + \(\dfrac{2021}{2}\)) + (1 + \(\dfrac{2020}{3}\)) + ... + ( 1 + \(\dfrac{1}{2022}\)) + 1
TS = \(\dfrac{2023}{2}\) + \(\dfrac{2023}{3}\) +...+ \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2023}\)
TS = 2023.(\(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) +...+ \(\dfrac{1}{2023}\))
A = \(\dfrac{2023.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\right)}{\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\right)}\)
A = 2023
LT
1
11 tháng 9 2023
Ta có :
\(\dfrac{10^{2023}}{10^{2024}}=\dfrac{10^{2022}}{10^{2023}}\)
mà \(\dfrac{10^{2023}}{10^{2024}}>\dfrac{10^{2023}-3}{10^{2024}-3}\)
\(\dfrac{10^{2022}}{10^{2023}}< \dfrac{10^{2022}+1}{10^{2023}+1}\)
\(\Rightarrow\dfrac{10^{2023}-3}{10^{2024}-3}< \dfrac{10^{2022}+1}{10^{2023}+1}\)
NH
0
HN
0
T
23 tháng 8 2023
\(B=\dfrac{\dfrac{2ab}{3}-\dfrac{3ab}{2}}{-\dfrac{5bb}{6}}\)
\(=\dfrac{\dfrac{4ab}{6}-\dfrac{9ab}{6}}{-\dfrac{5bb}{6}}\)
\(=\dfrac{-\dfrac{5ab}{6}}{-\dfrac{5bb}{6}}=\dfrac{ab.\dfrac{5}{6}}{bb.\dfrac{5}{6}}\)
\(=\dfrac{ab}{bb}=\dfrac{a}{b}\)
Với \(a=\dfrac{2021}{2022};b=\dfrac{2023}{2022}\), ta được:
\(B=\dfrac{2021}{2022}:\dfrac{2023}{2022}=\dfrac{2021}{2022}.\dfrac{2022}{2023}=\dfrac{2021}{2023}\)
OC
0
ND
1
AH
Akai Haruma
Giáo viên
25 tháng 3 2023
Lời giải:
$A=\frac{1}{4}(1-3+3^2-3^3+...+3^{2022}-3^{2023})$
$3A=\frac{1}{4}(3-3^2+3^3-3^4+....+3^{2023}-3^{2024})$
$3A+A=\frac{1}{4}(3-3^2+3^3-3^4+....+3^{2023}-3^{2024}+1-3+3^2-3^3+...+3^{2022}-3^{2023})$
$4A=\frac{1}{4}(1-3^{2024})$
$A=\frac{1}{16}(1-3^{2024})$
NH
23 tháng 8 2023
\(A=\dfrac{3\cdot\dfrac{a}{b}-\dfrac{-a}{b}}{-\dfrac{-5a}{b}+\dfrac{4a}{b}}\\ =\left(\dfrac{3a}{b}+\dfrac{a}{b}\right):\left(\dfrac{5a}{b}+\dfrac{4a}{b}\right)\\ =\dfrac{4a}{b}:\dfrac{9a}{b}\\ =\dfrac{4a}{b}\cdot\dfrac{b}{9a}\\ =\dfrac{4}{9}\)
Vậy `a=2021/2022` ; `b=2023/2022` thì `A=4/9`
Lời giải:
$A=1-3+3^2-3^3+...+3^{2022}-\frac{3^{2023}}{4}$
$3A=3-3^2+3^3-3^4+...+3^{2023}-\frac{3^{2024}}{4}$
$\Rightarrow A+3A=1+3^{2023}-\frac{3^{2023}}{4}-\frac{3^{2024}}{4}$
$\Rightarrow 4A=1$
$\Rightarrow A=\frac{1}{4}$