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Lời giải:
Ta có:
$(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=2023.\frac{2024}{2023}$
$\Leftrightarrow 1+\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+1+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}+1=2024$
$\Leftrightarrow 3+\frac{x+z}{y}+\frac{y+z}{x}+\frac{x+y}{z}=2024$
$\Leftrightarrow 3+B=2024$
$\Leftrightarrow B=2021$
`@` `\text {Ans}`
`\downarrow`
`(-2023) \times 33+2023 \times (-68)+2023`
`= 2023 \times (-33 - 68 + 1)`
`= 2023 \times (-101 + 1)`
`= 2023 \times (-100)`
`= -202300`
-100 thì nhân với 2023 phải là -202300 mới đúng chứ ạ
=> 4S = 1 + 2/4 + 3/4^2 +...+ 2023/4^2022
=> 4S-S = 1 + (2/4-1/4) + (3/4^2 - 2/4^2) +...+ (2023/4^2022 - 2022/4^2022) - 2023/4^2023
=> 3S = 1 + 1/4 + 1/4^2 +...+ 1/4^2022 - 2023/4^2023
=> 12S = 4 + 1 + 1/4 +... + 1/4^2021 - 2023/4^2022
=> 12S - 3S = 4 + (1-1) + (1/4-1/4) +... + (1/4^2021 - 1/4^2021) - 1/4^2022 - 2023/4^2022 + 2023/4^2023
=> 9S = 4 - 1/4^2022 - 2023/4^2022 + 2023/4^2023
= 4- 2024/4^2022 + 2023/4^2023
Do 2024/4^2022 > 2024/4^2023 > 2023/4^2023 nên - 2024/4^2022 + 2023/4^2023 < 0
=> 9S < 4 < 9/2
=> S < 1/2 (đpcm)
Ta có S = \(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2023}{4^{2023}}\)
4S = \(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2023}{4^{2022}}\)
4S - S = ( \(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2023}{4^{2022}}\) ) - ( \(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2023}{4^{2023}}\))
3S = 1 + \(\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2022}}-\dfrac{2023}{4^{2023}}\)
Đặt A = 1 + \(\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2022}}\)
4A = 4 + 1 + \(\dfrac{1}{4}+...+\dfrac{1}{4^{2021}}\)
4A - A = ( 4 + 1 + \(\dfrac{1}{4}+...+\dfrac{1}{4^{2021}}\)) - ( 1 + \(\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2022}}\))
3A = 4 - \(\dfrac{1}{4^{2022}}\)
A = ( 4 - \(\dfrac{1}{4^{2022}}\)) : 3 = \(\dfrac{4}{3}-\dfrac{1}{4^{2022}\cdot3}\)
⇒ 3S = \(\dfrac{4}{3}-\dfrac{1}{4^{2022}\cdot3}\) - \(\dfrac{2023}{4^{2023}}\)
S = ( \(\dfrac{4}{3}-\dfrac{1}{4^{2022}\cdot3}\) - \(\dfrac{2023}{4^{2023}}\)) : 3 = \(\dfrac{4}{9}-\dfrac{1}{4^{2022}\cdot3^2}-\dfrac{1}{4^{2023}\cdot3}< \dfrac{4}{9}< \dfrac{1}{2}\)
Vậy S < \(\dfrac{1}{2}\)
Xét VT : x+3x+5x+7x+......+2023x
Số hạng của dãy số trên là : \(\dfrac{2023-1}{2}+1=1012\left(sốhạng\right)\)
Tổng số của dãy số trên là : \(\dfrac{\left(2023x+x\right).1012}{2}\text{=}1012x.1012\)
Do đó : ta có :
\(1012x.1012\text{=}2023.2024\)
\(1012x\text{=}4046\)
\(x\text{=}\dfrac{2023}{506}\)
VT = x + 3x + 5x + 7x +... + 2023x = [(2023 - 1):2 +1] . (2023+1)x = 1012. 2024x = 2048288x
VP= 2023 . 2024= 4094552
VT=VP <=> 2048288x =4094552
<=>\(x\approx2\)
Có `xyz=2023=>2023=xyz`
Thay vào ta có :
\(\dfrac{xyz\cdot x}{xy+xyz\cdot x+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{xz+z+1}=1\\ \dfrac{x^2yz}{xy\left(1+xz+z\right)}+\dfrac{y}{y\left(z+1+xz\right)}+\dfrac{z}{xz+z+1}=1\\ \dfrac{xz}{1+xz+z}+\dfrac{1}{z+1+xz}+\dfrac{z}{xz+z+1}=1\\ \dfrac{xz+1+z}{1+xz+z}=1\left(dpcm\right)\)