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A = 1 + 2 + 22 + 23 + 24 + ..... + 22021
2A = 2 + 22 + 23 + 24 + 25 + ..... + 22022
2A - A = ( 2 + 22 + 23 + 24 + 25 + ..... + 22022 ) - ( 1 + 2 + 22 + 23 + 24 + ..... + 22021 )
A = 22022 - 1
\(3n-2\inƯ\left(15\right)\) \(=\left\{1;-1;3;-3;5;-5;15;-15\right\}.\)
\(\Leftrightarrow n\in\left\{1;\dfrac{1}{3};\dfrac{5}{3};\dfrac{-1}{3};\dfrac{7}{3};-1;\dfrac{17}{3};\dfrac{-13}{3}\right\}.\)
Mà \(n\ne\dfrac{2}{3};n\in Z.\)
\(\Rightarrow n\in\left\{1;-1\right\}.\)
\(f,=\left(5^2+3\right):7=28:7=4\\ g,=7^2-9+8\cdot25=49-9+200=240\\ h,=600+72+18=690\\ i,=5^2+5-20=10\\ j,=45-28+83=100\)
\(2A=\frac{4}{1.5}+\frac{6}{5.11}+\frac{8}{11.19}+\frac{10}{19.29}+\frac{12}{29.41}\)
\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{19}+...+\frac{1}{29}-\frac{1}{41}=1-\frac{1}{41}=\frac{40}{41}\)
\(\Rightarrow A=\frac{20}{21}\)
\(3B=\frac{3}{1.4}+\frac{6}{4.10}+\frac{9}{10.19}+\frac{12}{19.31}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{10}+\frac{1}{10}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}\)
\(=1-\frac{1}{31}=\frac{30}{31}\)
\(\Rightarrow B=\frac{10}{31}=\frac{20}{62}<\frac{20}{41}\)
Do đó $A>B$
Ta có: \(A=\dfrac{2}{1.5}+\dfrac{3}{5.11}+\dfrac{4}{11.19}+\dfrac{5}{19.29}+\dfrac{6}{29.41}\)
\(2A=1-\dfrac{1}{5}+\dfrac{1}{5}+...+\dfrac{1}{29}-\dfrac{1}{41}\)
\(2A=1-\dfrac{1}{41}=\dfrac{40}{41}\)
\(A=\dfrac{20}{41}\)
Lại có: \(B=\dfrac{1}{1.4}+\dfrac{2}{4.10}+\dfrac{3}{10.19}+\dfrac{4}{19.31}\)
\(3B=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}\)
\(3B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+...+\dfrac{1}{19}-\dfrac{1}{31}\)
\(3B=1-\dfrac{1}{31}=\dfrac{30}{31}\)
\(B=\dfrac{10}{31}\)
Vì \(\dfrac{20}{41}>\dfrac{10}{31}\) nên...
(143+35)x25+75x(39+139)=178x25+75x178=178x(25+75)=178x100=17800
143 . 25 + 25 . 35 + 75 . 39 + 75 . 139
= [ 25.( 143 + 35 ) ] + [ 75 . ( 39 + 139 ) ]
= 25 . 178 + 75 . 178
= ( 25 + 75 ) . 178
= 100 . 178
= 17800 .
Hok tốt :))
`Answer:`
\(B=\frac{101}{120}+\frac{1}{2.6}+\frac{1}{4.9}+\frac{1}{6.12}+...+\frac{1}{36.57}+\frac{1}{38.60}\)
\(=\frac{101}{120}+\frac{1}{12}+\frac{1}{36}+\frac{1}{72}+...+\frac{1}{2052}+\frac{1}{2280}\)
\(=\frac{101}{120}+\frac{1}{6}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{342}+\frac{1}{380}\right)\)
\(=\frac{101}{120}+\frac{1}{6}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(=\frac{101}{120}+\frac{1}{6}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=\frac{101}{120}+\frac{1}{6}.\left(1-\frac{1}{20}\right)\)
\(=\frac{101}{120}+\frac{1}{6}.\frac{19}{20}\)
\(=\frac{101}{120}+\frac{19}{120}\)
\(=1\)
`->` Chọn D.