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DH
1
KV
19 tháng 10 2023
a) 5.3²⁰²³ = 50.3²⁰²³ - 5.9ˣ
5.9ˣ = 50.3²⁰²³ - 5.3²⁰²³
5.(3²)ˣ = 5.3²⁰²³.(10 - 1)
5.(3²)ˣ = 5.3²⁰²³.9
3²ˣ = 3²⁰²³.3²
3²ˣ = 3²⁰²⁵
2x = 2025
x = 2025/2
b) 2.3ˣ + 5.3ˣ⁺¹ = 153
3ˣ.(2 + 5.3) = 153
3ˣ.17 = 153
3ˣ = 153/17
3ˣ = 9
3ˣ = 3²
x = 2
ND
1
HK
Cho \(A=\dfrac{2023^{30}+5}{2023^{31}+5}\) và \(B=\dfrac{2023^{31}+5}{2023^{32}+5}\). So sánh A và B
2
17 tháng 4 2023
Áp dụng tính chất : Nếu \(\dfrac{a}{b}< 1\) thì \(\dfrac{a}{b}< \dfrac{a+n}{b+n}\) ( a; b; n ϵ N , b; n ≠ 0 )
Ta có \(\dfrac{2023^{31}+5}{2023^{32}+5}< 1\)
⇒ \(B=\dfrac{2023^{31}+5}{2023^{32}+5}< \dfrac{2023^{31}+5+2018}{2023^{32}+5+2018}=\dfrac{2023^{31}+2023}{2023^{32}+2023}=\dfrac{2023\left(2023^{30}+1\right)}{2023\left(2023^{31}+1\right)}=\dfrac{2023^{30}+1}{2023^{31}+1}=A\)Vậy A > B
XO
17 tháng 4 2023
Ta có 2023A = \(\dfrac{2023.\left(2023^{30}+5\right)}{2023^{31}+5}=\dfrac{2023^{31}+5.2023}{2023^{31}+5}\)
\(=1+\dfrac{2022.5}{2023^{31}+5}\)
Lại có 2023B = \(\dfrac{2023.\left(2023^{31}+5\right)}{2023^{32}+5}=\dfrac{2023^{32}+2023.5}{2023^{32}+5}\)
\(=1+\dfrac{2022.5}{2023^{32}+5}\)
Dễ thấy 202331 + 5 < 202332 + 5
\(\Leftrightarrow\dfrac{2022.5}{2023^{31}+5}>\dfrac{2022.5}{2023^{32}+5}\)
\(\Leftrightarrow1+\dfrac{2022.5}{2023^{31}+5}>1+\dfrac{2022.5}{2023^{32}>5}\)
\(\Leftrightarrow2023A>2023B\Leftrightarrow A>B\)
TL
1
23 tháng 2 2023
\(2023A=\dfrac{2023^{31}+4046}{2023^{31}+2}=1+\dfrac{4044}{2023^{31}+2}\)
\(2023B=\dfrac{2023^{32}+4046}{2023^{32}+2}=1+\dfrac{4044}{2023^{32}+2}\)
mà 2023^31+2<2023^32+2
nên A>B
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
PN
1
\(A=\dfrac{2023}{1\cdot2}+\dfrac{2023}{2\cdot3}+...+\dfrac{2023}{2022\cdot2023}\)
\(=2023\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2022\cdot2023}\right)\)
\(=2023\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)
\(=2023\left(1-\dfrac{1}{2023}\right)=2023\cdot\dfrac{2022}{2023}=2022\)
\(A=\dfrac{2023}{1.2}+\dfrac{2023}{2.3}+\dfrac{2023}{3.4}+...+\dfrac{2023}{2022.2023}\)
\(A=\dfrac{2023}{1}.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2022.2023}\right)\)
\(A=\dfrac{2023}{1}.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)
\(A=\dfrac{2023}{1}.\left(1-\dfrac{1}{2023}\right)\)
\(A=\dfrac{2023}{1}.\dfrac{2022}{2023}\)
\(A=1.2022\)
\(A=2022\)
Vậy \(A=2022\)