a) \(1+2+3+...+48=\dfrac{\left(48+1\right)\left(\dfrac{48-1}{1}+1\right)}{2}=1176\)

b) \(2+4+6+...+212=\dfrac{\left(212+2\right)\left(\dfrac{212-2}{2}+1\right)}{2}=11342\)

b: Ta có: \(\left(2+4+6+...+100\right)\cdot\left(36\cdot333-108\cdot111\right)\)

\(=\left(2+4+6+...+100\right)\cdot36\cdot111\cdot\left(3-3\right)\)

=0

\(f,=\left(5^2+3\right):7=28:7=4\\ g,=7^2-9+8\cdot25=49-9+200=240\\ h,=600+72+18=690\\ i,=5^2+5-20=10\\ j,=45-28+83=100\)

\(2A=\frac{4}{1.5}+\frac{6}{5.11}+\frac{8}{11.19}+\frac{10}{19.29}+\frac{12}{29.41}\)

\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{19}+...+\frac{1}{29}-\frac{1}{41}=1-\frac{1}{41}=\frac{40}{41}\)

\(\Rightarrow A=\frac{20}{21}\)

\(3B=\frac{3}{1.4}+\frac{6}{4.10}+\frac{9}{10.19}+\frac{12}{19.31}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{10}+\frac{1}{10}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}\)

\(=1-\frac{1}{31}=\frac{30}{31}\)

\(\Rightarrow B=\frac{10}{31}=\frac{20}{62}<\frac{20}{41}\)

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...

đk : \(n\ne-\dfrac{1}{3}\)

gọi d là ƯCLN(18n+3,21n+7)

ta có 18n+3chia hết cho d

          21n+7 chia hết cho d

⇔21n+7-18n-3 chia hết cho d

⇔126n+42-126n-21 chia hết cho d 

21 chia hết cho d

⇒d∈Ư(21)=1;3;7;21

n ≠ 3k-1;3k-3;3k-7;3k-21

umm..đề bài đâu bn?

Đề bài pay mất r bn<3

\(3n-2\inƯ\left(15\right)\) \(=\left\{1;-1;3;-3;5;-5;15;-15\right\}.\)

\(\Leftrightarrow n\in\left\{1;\dfrac{1}{3};\dfrac{5}{3};\dfrac{-1}{3};\dfrac{7}{3};-1;\dfrac{17}{3};\dfrac{-13}{3}\right\}.\)

Mà \(n\ne\dfrac{2}{3};n\in Z.\)

\(\Rightarrow n\in\left\{1;-1\right\}.\)

\(c,\overline{735a2b}⋮̸2;⋮5\Rightarrow b=5\\ \Rightarrow\overline{735a25}⋮9\\ \Rightarrow7+3+5+a+2+5⋮9\\ \Rightarrow22+a⋮9\\ \Rightarrow a=5\\ \Rightarrow\overline{735a2b}=735525\\ b,\overline{5a27b}⋮2;⋮5\Rightarrow b=0\\ \Rightarrow\overline{5a270}⋮9\\ \Rightarrow5+a+2+7+0⋮9\\ \Rightarrow14+a⋮9\\ \Rightarrow a=4\\ \Rightarrow\overline{5a27b}=54270\)

c) \(735525\)
d) \(54270\)