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(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = (\(\dfrac{2021}{2}+1\))+(\(\dfrac{2020}{3}+1\))+....+(\(\dfrac{1}{2022}+1\))
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = \(\dfrac{2023}{2}\)+\(\dfrac{2023}{3}\)+....+ \(\dfrac{2023}{2022}\)
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = 2023.( \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\))
vậy x= 2023
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{3}{4}B=\dfrac{3}{4}-\left(\dfrac{3}{4}\right)^2+\left(\dfrac{3}{4}\right)^3-....-\left(\dfrac{3}{4}\right)^{2024}+\left(\dfrac{3}{4}\right)^{2025}\)
=>\(\dfrac{7}{4}B=\left(\dfrac{3}{4}\right)^{2025}+1\)
=>\(B\cdot\dfrac{7}{4}=\dfrac{3^{2025}+4^{2025}}{4^{2025}}\)
=>\(B=\dfrac{3^{2025}+4^{2025}}{4^{2024}\cdot7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{2024^{2023}+1}{2024^{2024}+1}\)
\(2024A=\dfrac{2024^{2024}+2024}{2024^{2024}+1}=\dfrac{\left(2024^{2024}+1\right)+2023}{2024^{2024}+1}=\dfrac{2024^{2024}+1}{2024^{2024}+1}+\dfrac{2023}{2024^{2024}+1}=1+\dfrac{2023}{2024^{2024}+1}\)
\(B=\dfrac{2024^{2022}+1}{2024^{2023}+1}\)
\(2024B=\dfrac{2024^{2023}+2024}{2024^{2023}+1}=\dfrac{\left(2024^{2023}+1\right)+2023}{2024^{2023}+1}=\dfrac{2024^{2023}+1}{2024^{2023}+1}+\dfrac{2023}{2024^{2023}+1}=1+\dfrac{2023}{2024^{2023}+1}\)
Vì \(2024>2023=>2024^{2024}>2024^{2023}\)
\(=>2024^{2024}+1>2024^{2023}+1\)
\(=>\dfrac{2023}{2024^{2023}+1}>\dfrac{2023}{2024^{2024}+1}\)
\(=>A< B\)
\(#PaooNqoccc\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`#040911`
`a)`
`3 1/3 x + 16 3/4 = -13,25`
`=> 3 1/3 x = -13,25 - 16 3/4`
`=> 3 1/3 x = -30`
`=> x = -30 \div 3 1/3`
`=> x =-9`
Vậy, `x = -9`
`b)`
`3 2/7*x - 1/8 = 2 3/4`
`=> 3 2/7x = 2 3/4 + 1/8`
`=> 3 2/7x = 23/8`
`=> x = 23/8 \div 3 2/7`
`=> x = 7/8`
Vậy, `x = 7/8`
`c)`
`x \div 4 1/3 = -2,5`
`=> x = -2,5 * 4 1/3`
`=> x = -65/6`
Vậy, `x = -65/6`
`d)`
`( (3x)/7 + 1) \div (-4) = (-1)/28`
`=> (3x)/7 +1 = (-1)/28 * (-4)`
`=> (3x)/7 + 1 = 1/7`
`=> (3x)/7 = 1/7 - 1`
`=> (3x)/7 = -6/7`
`=> 3x = -6`
`=> x = -6 \div 3`
`=> x = -2`
Vậy, `x = -2.`
a
=>10/3 . x + 16 + 3/4 = -13,25
=>10/3 x + 3/4 = -29,25
=>10/3 x = -30
=>x=-30 : 10/3
=>x=-30 . 3/10
=>x=-9
b.
=>23/7 x - 1/8 = = 11/4
=>23/7 x = 11/4 + 1/8
=>23/7 x= 22/8 + 1/8
=>23/7 x= 23/8
=>x=23/8 : 23/7
=>x=23/8 . 7/23
=>x=7/8
c.
=>x : 13/3 =-5/2
=>x=-5/2 . 13/3
=>x=-65/6
d.
=>3x/7 +1 = (-1/28) . (-4)
=>3x/7 + 1 = 1/7
=>3x/7 = -6/7
=>3x=-6
=>x=-2
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=\dfrac{2^{2024}-3}{2^{2023}-1}=\dfrac{2.2^{2023}-2-1}{2^{2023}-1}=\dfrac{2\left(2^{2023}-1\right)-1}{2^{2023}-1}=2-\dfrac{1}{2^{2023}-1}\)
\(D=\dfrac{2^{2023}-3}{2^{2022}-1}=\dfrac{2.2^{2022}-2-1}{2^{2022}-1}=\dfrac{2\left(2^{2022}-1\right)-1}{2^{2022}-1}=2-\dfrac{1}{2^{2022}-1}\)
Ta có
\(2^{2023}>2^{2022}\Rightarrow2^{2023}-1>2^{2022}-1\)
\(\Rightarrow\dfrac{1}{2^{2023}-1}< \dfrac{1}{2^{2022}-1}\Rightarrow2-\dfrac{1}{2^{2023}-1}>2-\dfrac{1}{2^{2022}-1}\)
\(\Rightarrow C>D\)