so sánh A và B biết
A=\(\dfrac{2^{2021}+1}{2^{2021}}\) và B=\(\dfrac{2^{2021}+2}{2^{2021}+1}\)
Giúp mk với đc không mk đang gấp
cảm ơn nhiều
So sánh: A = \(\dfrac{2021^{10}-1}{2021^9-1}\)và B = \(\dfrac{2021^{11}-1}{2021^{10}-1}\) làm ơn hãy giúp tôii đi màaaa
\(A=\dfrac{2021^{10}-2021+2020}{2021^9-1}\\ =\dfrac{2021\left(2021^9-1\right)+2020}{2021^9-1}\\ =2021+\dfrac{2020}{2021^9-1}\\ B=\dfrac{2021^{11}-1}{2021^{10}-1}=2021+\dfrac{2020}{2021^{10}-1}\)
Ta có:
\(2021^9-1< 2021^{10}-1\\ \Rightarrow\dfrac{2020}{2021^9-1}>\dfrac{2020}{2021^{10}-1}\)
Do đó A > B.
So sánh:
A = \(\dfrac{2^{2020}-1}{2^{2021}-1}\) và B = \(\dfrac{2^{2021}-1}{2^{2022}-1}\)
\(2.A=\frac{2^{2021}-2}{2^{2021}-1}=1-\frac{1}{2^{2021}-1}\)
\(2B=\frac{2^{2022}-2}{2^{2022}-1}=1-\frac{1}{2^{2022}-1}\)
dó \(\frac{1}{2^{2022}-1}< \frac{1}{2^{2021}-1}\Rightarrow1-\frac{1}{2^{2022}-1}>1-\frac{1}{2^{2021}-1}\Rightarrow A< B\)
HT
Không tính kết quả hãy so sánh:
A) 2008×2010 và 2009×2009
B) 2020×2022 và 2021×2021
Giúp mk nhé!!
Mình cảm ơn nhiều !! 💗💗
Vì sao bằng nhau ạ?? Xin hãy cho giải thích kỹ ak
1. So sánh
a) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\) và B= \(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{13}{60}\)
b) \(C=\dfrac{2019}{2021}+\dfrac{2021}{2022}\) và \(D=\dfrac{2020+2022}{2019+2021}.\dfrac{3}{2}\)
a) Ta có:
2A=2.(12+122+123+...+122020+122021)2�=2.12+122+123+...+122 020+122 021
2A=1+12+122+123+...+122019+1220202�=1+12+122+123+...+122 019+122 020
Suy ra: 2A−A=(1+12+122+123+...+122019+122020)2�−�=1+12+122+123+...+122 019+122 020
−(12+122+123+...+122020+122021)−12+122+123+...+122 020+122 021
Do đó A=1−122021<1�=1−122021<1.
Lại có B=13+14+15+1360=20+15+12+1360=6060=1�=13+14+15+1360=20+15+12+1360=6060=1.
Vậy A < B.
ta có B= \(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+\dfrac{4}{4^4}+.....+\dfrac{2021}{4^{2021}}\)
so sánh B với \(\dfrac{1}{2}\)
giúp mik với mình cần gấp mai thi rồi
Lời giải:
\(B=\frac{1}{4}+\frac{2}{4^2}+\frac{3}{4^3}+....+\frac{2021}{4^{2021}}\)
\(4B=1+\frac{2}{4}+\frac{3}{4^2}+...+\frac{2021}{4^{2020}}\)
\(4B-B=1+\frac{1}{4}+\frac{1}{4^2}+...+\frac{1}{4^{2020}}-\frac{2021}{4^{2021}}\)
\(3B=1+\frac{1}{4}+\frac{1}{4^2}+....+\frac{1}{4^{2020}}-\frac{2021}{4^{2021}}\)
\(12B=4+1+\frac{1}{4}+...+\frac{1}{4^{2019}}-\frac{2021}{4^{2020}}\)
\(9B=4-\frac{6067}{4^{2021}}<4\Rightarrow B< \frac{4}{9}< \frac{1}{2}\)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2021^2}\). So sánh A và \(\dfrac{2020}{2021}\)
a) tìm x :\(\dfrac{2}{1.4}x+\dfrac{2}{4.7}x+\dfrac{2}{7.10}x+....+\dfrac{2}{31.344}x=10\)
b)so sánh hai phân số sau : A=\(\dfrac{6^{2020}+1}{6^{2021}+1}\)và B=\(\dfrac{6^{\text{2021}}+1}{\text{6}^{\text{2022}}+1}\)
ét o ét giúp với ạ
S=\(\dfrac{2}{2021+1}+\dfrac{2^2}{2021^2+1}+\dfrac{2^3}{2021^{2^2}}+...+\dfrac{2^{n+1}}{2021^{2^n}+1}+...+\dfrac{2^{2021}}{2021^{2^{2020}}+1}\)so sánh S với \(\dfrac{1}{1010}\)
\(S=\dfrac{2}{2021+1}+\dfrac{2^2}{2021^2+1}+\dfrac{2^3}{2021^{2^2}+1}+...+\dfrac{2^{n+1}}{2021^{2^n}+1}+...+\dfrac{2^{2021}}{2021^{2^{2020}}+1}\)
So sánh S với \(\dfrac{1}{1010}\)