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NL
Nguyễn Lê Phước Thịnh
CTVHS
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
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 \(M< N\)
ND
2
NM
0
NL
Nguyễn Lê Phước Thịnh
CTVHS
6 tháng 10 2025
Sửa đề: Tìm x
a: \(x\left(6-x\right)^{2023}=\left(6-x\right)^{2023}\)
=>\(x\left(6-x\right)^{2023}-\left(6-x\right)^{2023}=0\)
=>\(\left(6-x\right)^{2023}\left(x-1\right)=0\)
=>\(\left[\begin{array}{l}6-x=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=6\\ x=1\end{array}\right.\)
b: \(5^{x}+5^{x+2}=650\)
=>\(5^{x}+5^{x}\cdot25=650\)
=>\(5^{x}\left(1+25\right)=650\)
=>\(5^{x}=\frac{650}{26}=25=5^2\)
=>x=2
c: \(2^{x+2}-2^{x}=96\)
=>\(2^{x}\cdot2^2-2^{x}=96\)
=>\(2^{x}\left(2^2-1\right)=96\)
=>\(2^{x}\cdot3=96\)
=>\(2^{x}=\frac{96}{3}=32=2^5\)
=>x=5
d: \(10^{x}:5^{y}=20^{y}\)
=>\(10^{x}=20^{y}\cdot5^{y}=100^{y}=\left(10\right)^{2y}\)
=>x=2y
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 \(\dfrac{9}{10^{2023}+1}< \dfrac{9}{10^{2022}+1}\)
\(\Rightarrow A>B\)
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
\(\Leftrightarrow\dfrac{2022.5}{2023^{31}+5}>\dfrac{2022.5}{2023^{32}+5}\)
\(\Leftrightarrow1+\dfrac{2022.5}{2023^{31}+5}>1+\dfrac{2022.5}{2023^{32}>5}\)
\(\Leftrightarrow2023A>2023B\Leftrightarrow A>B\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
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
\(2023A=\dfrac{2023^{12}+2\cdot2023}{2023^{12}+2}=1+\dfrac{4044}{2023^{12}+2}\)
\(2023B=\dfrac{2023^{11}+4046}{2023^{11}+2}=1+\dfrac{4044}{2023^{11}+2}\)
Ta có: \(2023^{12}+2>2023^{11}+2\)
=>\(\dfrac{4044}{2023^{12}+2}< \dfrac{4044}{2023^{11}+2}\)
=>\(\dfrac{4044}{2023^{12}+2}+1< \dfrac{4044}{2023^{11}+2}+1\)
=>2023A<2023B
=>A<B
2023A=202312+2202312+2⋅2023=1+202312+24044
\(2023 B = \frac{202 3^{11} + 4046}{202 3^{11} + 2} = 1 + \frac{4044}{202 3^{11} + 2}\)
Ta có: \(202 3^{12} + 2 > 202 3^{11} + 2\)
=>\(\frac{4044}{202 3^{12} + 2} < \frac{4044}{202 3^{11} + 2}\)
=>\(\frac{4044}{202 3^{12} + 2} + 1 < \frac{4044}{202 3^{11} + 2} + 1\)
=>2023A<2023B
=>A<B