M=3/1+2 + 3/1+2+3 + 3/1+2+3+4 + ...+ 3/1+2+...+2022
so sánh M với 10/3
mọi người giúp e vs , e đang cần gấp
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HH
2
Những câu hỏi liên quan
9 tháng 1
\(M=\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+...+2022}\)
\(=\dfrac{3}{\dfrac{2\left(2+1\right)}{2}}+\dfrac{3}{\dfrac{3\left(3+1\right)}{2}}+...+\dfrac{3}{\dfrac{2022\left(2022+1\right)}{2}}\)
\(=\dfrac{6}{2\left(2+1\right)}+\dfrac{6}{3\left(3+1\right)}+...+\dfrac{6}{2022\cdot2023}\)
\(=\dfrac{6}{2\cdot3}+\dfrac{6}{3\cdot4}+...+\dfrac{6}{2022\cdot2023}\)
\(=6\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2022\cdot2023}\right)\)
\(=6\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)
\(=6\cdot\left(\dfrac{1}{2}-\dfrac{1}{2023}\right)=6\cdot\dfrac{2021}{4046}=\dfrac{12126}{4046}< 3\)
mà \(3< \dfrac{10}{3}\)
nên \(M< \dfrac{10}{3}\)
27 tháng 10 2023
\(2^{35}< 2^{36}=8^{12}\)
\(3^{24}=\left(3^2\right)^{12}=9^{12}\)
\(8^{12}< 9^{12}\)
=>\(2^{36}< 3^{24}\)
=>\(2^{35}< 3^{24}\)
XT
2
16 tháng 10 2021
\(4\left(a^3+b^3\right)-6\left(a^2+b^2\right)\)
\(=4\left(a+b\right)^3-12ab\left(a+b\right)-6\left(a+b\right)^2+12ab\)
\(=4-6-12ab+12ab\)
=-2
14 tháng 6 2023
M=1/4(4/1*5+8/5*13+12/13*15+16/25*41)
=1/4(1-1/5+1/5-1/13+...+1/25-1/41)
=1/4*40/41=10/41
N=1/3(6/1*7+9/7*16+...+18/43*61)
=1/3(1-1/7+...+1/43-1/61)
=1/3*60/61=20/41
=>M<N
H9
HT.Phong (9A5)
CTVHS
26 tháng 3 2023
\(\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(8^2-576:3^2\right)\)
\(=\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(64-576:3^2\right)\)
\(=\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(64-64\right)\)
\(=\left(1^1+2^2+3^3+4^4+2022^{2022}\right).0\)
\(=0\)
AH
Akai Haruma
Giáo viên
28 tháng 1
Câu 1:
$B=\frac{10}{1.3}+\frac{10}{3.5}+\frac{10}{5.7}+...+\frac{10}{101.103}$
$B=5(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{101.103})$
$=5(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{103-101}{101.103})$
$=5(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103})$
$=5(1-\frac{1}{103})=5.\frac{102}{103}=\frac{510}{103}$
AH
Akai Haruma
Giáo viên
28 tháng 1
Câu 2:
\(C=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+....+\frac{1}{2022.2024}\\ =\frac{1}{2}\left[\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{2022.2024}\right]\)
\(=\frac{1}{2}\left[\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+....+\frac{2024-2022}{2022.2024}\right]\)
\(=\frac{1}{2}(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2022}-\frac{1}{2024})\\ =\frac{1}{2}(\frac{1}{2}-\frac{1}{2024})=\frac{1011}{4048}\)
mọi người ơi giúp em vs ạ , e đang rất cần
\(1+2+...+n=\dfrac{\left(\dfrac{n-1}{1}+1\right).\left(n+1\right)}{2}=\dfrac{n\left(n+1\right)}{2}\)
\(M=\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+...+2022}\)
\(=3\left(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+...+2022}\right)\)
\(=3\left(\dfrac{1}{\dfrac{2.\left(2+1\right)}{2}}+\dfrac{1}{\dfrac{3.\left(3+1\right)}{2}}+...+\dfrac{1}{\dfrac{2022.\left(2022+1\right)}{2}}\right)\)
\(=3\left(\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{2022.2023}\right)\)
\(=3.2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)\)
\(=6.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)
\(=6.\left(\dfrac{1}{2}-\dfrac{1}{2023}\right)\)
\(=6.\dfrac{2021}{4046}=3.\dfrac{2021}{2023}=\dfrac{6063}{2023}=\dfrac{18189}{6069}\)
\(\dfrac{10}{3}=\dfrac{20230}{6069}>\dfrac{18189}{6069}=M\)