NK
1
K
Khách
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ND
1
TP
0
24 tháng 7 2023
a: \(B=\dfrac{154}{155+156}+\dfrac{155}{155+156}\)
\(\dfrac{154}{155}>\dfrac{154}{155+156}\)
\(\dfrac{155}{156}>\dfrac{155}{155+156}\)
=>154/155+155/156>(154+155)/(155+156)
=>A>B
b: \(C=\dfrac{2021+2022+2023}{2022+2023+2024}=\dfrac{2021}{6069}+\dfrac{2022}{6069}+\dfrac{2023}{6069}\)
2021/2022>2021/6069
2022/2023>2022/2069
2023/2024>2023/6069
=>D>C
VN
Võ Ngọc Phương
VIP
13 tháng 9 2023
b) \(M=\dfrac{10^{2023}+1}{10^{2024}+1}< 1\) ( Vì tử < mẫu )
Ta có: \(M=\dfrac{10^{2023}+1}{10^{2024}+1}< \dfrac{10^{2023}+1+9}{10^{2024}+1+9}=\dfrac{10^{2023}+10}{10^{2024}+10}=\dfrac{10.\left(10^{2022}+1\right)}{10.\left(10^{2023}+1\right)}=\dfrac{10^{2022}+1}{10^{2023}+1}=N\)
Vì \(\dfrac{10^{2023}+1}{10^{2024}+1}< \dfrac{10^{2022}+1}{10^{2023}+1}\) nên \(M< N\)
NN
13 tháng 2 2023
\(A=\dfrac{2024^{2023}+1}{2024^{2024}+1}\)
\(2024A=\dfrac{2024^{2024}+2024}{2024^{2024}+1}=\dfrac{\left(2024^{2024}+1\right)+2023}{2024^{2024}+1}=\dfrac{2024^{2024}+1}{2024^{2024}+1}+\dfrac{2023}{2024^{2024}+1}=1+\dfrac{2023}{2024^{2024}+1}\)
\(B=\dfrac{2024^{2022}+1}{2024^{2023}+1}\)
\(2024B=\dfrac{2024^{2023}+2024}{2024^{2023}+1}=\dfrac{\left(2024^{2023}+1\right)+2023}{2024^{2023}+1}=\dfrac{2024^{2023}+1}{2024^{2023}+1}+\dfrac{2023}{2024^{2023}+1}=1+\dfrac{2023}{2024^{2023}+1}\)
Vì \(2024>2023=>2024^{2024}>2024^{2023}\)
\(=>2024^{2024}+1>2024^{2023}+1\)
\(=>\dfrac{2023}{2024^{2023}+1}>\dfrac{2023}{2024^{2024}+1}\)
\(=>A< B\)
\(#PaooNqoccc\)
V
2
MH
26 tháng 9 2023
\(A=\dfrac{10^{2024}+1}{10^{2023}+1}=\dfrac{10\left(10^{2023}+1\right)}{10^{2023}+1}-\dfrac{9}{10^{2023}+1}=1-\dfrac{9}{10^{2023}+1}\)
\(B=\dfrac{10^{2023}+1}{10^{2022}+1}=\dfrac{10\left(10^{2022}+1\right)}{10^{2022}+1}-\dfrac{9}{10^{2022}+1}=1-\dfrac{9}{10^{2022}+1}\)
Vì \(\dfrac{9}{10^{2023}+1}< \dfrac{9}{10^{2022}+1}\)
\(\Rightarrow A>B\)
NT
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
28 tháng 7 2023
\(C=\dfrac{2^{2024}-3}{2^{2023}-1}=\dfrac{2.2^{2023}-2-1}{2^{2023}-1}=\dfrac{2\left(2^{2023}-1\right)-1}{2^{2023}-1}=2-\dfrac{1}{2^{2023}-1}\)
\(D=\dfrac{2^{2023}-3}{2^{2022}-1}=\dfrac{2.2^{2022}-2-1}{2^{2022}-1}=\dfrac{2\left(2^{2022}-1\right)-1}{2^{2022}-1}=2-\dfrac{1}{2^{2022}-1}\)
Ta có
\(2^{2023}>2^{2022}\Rightarrow2^{2023}-1>2^{2022}-1\)
\(\Rightarrow\dfrac{1}{2^{2023}-1}< \dfrac{1}{2^{2022}-1}\Rightarrow2-\dfrac{1}{2^{2023}-1}>2-\dfrac{1}{2^{2022}-1}\)
\(\Rightarrow C>D\)
Ta có : \(N=2022.2024\)
\(N=\left(2023-1\right)\left(2023+1\right)\)
\(N=2023^2+2023-2023-1\)
\(N=2023^2-1\)
Mà : \(M=2023.2023=2023^2\)
\(\Rightarrow M>N\)