\(\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{7\times8}+....+\dfrac{1}{9\times10}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+.....+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=\dfrac{1}{3}-\dfrac{1}{10}\)

\(=\dfrac{7}{30}\)

84 = 2².7.3

Ư(84) = {1;84;2;42;4;21;14;6;28;3;12;7}

2.2.2.2.2........2 (có 2015 thừa số 2)

=> 2.2.2.2......2 = 2²⁰¹⁵ = 2²⁰¹². 2³ = (2⁴)⁵⁰³.8 = 16⁵⁰³.8

Ta biết : 16ⁿ (với n ϵ N*)

16ⁿ luôn có chữ số tận cùng là 6

=> 16⁵⁰³.8 = \(\overline{.....6}\cdot8=\overline{.......8}\)

Vậy 2.2.2.2.2........2 (có 2015 thừa số 2) có chữ số tận cùng là 8

4.4.4.4......4(có 2021 thừa số 4)

=> 4.4.4.4......4 = 4²⁰²¹ = 4²⁰²⁰.4 = (4²)¹⁰¹⁰.4 = 16¹⁰¹⁰.4

Ta biết: 16ⁿ ( với n ϵ N*)

16ⁿ luôn có chữ số tận cùng là 6

=> 16¹⁰¹⁰.4 = \(\overline{........6}\cdot4=\overline{.........24}\)

Vậy 4.4.4.....4.4.4 (có 2021 thừa số 4) có chữ số tận cùng là 4

\(\dfrac{1}{1\times2}+\dfrac{2}{2\times4}+\dfrac{3}{4\times7}+\dfrac{4}{7\times11}+....+\dfrac{11}{56\times67}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+....+\dfrac{1}{56}-\dfrac{1}{67}\)

\(=1-\dfrac{1}{67}\)

\(=\dfrac{1}{66}\)

\(a.\left[x\cdot\left(x+1\right)\right]:2=136\\ x\cdot\left(x+1\right)=136\cdot2\\ x\cdot\left(x+1\right)=2\cdot2\cdot2\cdot17\cdot2\\ x\cdot\left(x+1\right)=\left(2\cdot2\cdot2\cdot2\right)\cdot17\\ x\cdot\left(x+1\right)=16\cdot17\\ =>x=16\\ b.\left[x\cdot\left(x+1\right)\right]:2=300\\ x\cdot\left(x+1\right)=3\cdot2\cdot2\cdot5\cdot5\cdot2\\ x\cdot\left(x+1\right)=\left(3\cdot2\cdot2\cdot2\right)\cdot\left(5\cdot5\right)\\ x\cdot\left(x+1\right)=24\cdot25\\ =>x=24\\ c.\left[x\cdot\left(x+1\right)\right]:2=561\\ x\cdot\left(x+1\right)=2\cdot3\cdot11\cdot17\\ x\cdot\left(x+1\right)=\left(17\cdot2\right)\cdot\left(3\cdot11\right)\\ x\cdot\left(x+1\right)=34\cdot33\\ =>x=33\)

Em cảm ơn cô ạ

(3x + 52) ⋮ (x+7)

=> 3x + 52 - 3(x+7) ⋮ (x+7)

=> 3x + 52 - 3x - 21 ⋮ (x+7)

=> (3x - 3x) + (52 - 21) ⋮ (x+7)

=> 0 + 31 ⋮ (x+7)

=> 31 ⋮ (x+7)

=> x+7 ϵ Ư(31)

Ư(31) = {1;31;-31;-1}

=> x ϵ  {-6; 24 ; -38; -8}

\(\dfrac{19^5\cdot19^4}{19^7}\)

\(=\dfrac{19^9}{19^7}\\ =19^2\)

Ta biết : 2(a+b) = 2a + 2b

Tương tự như vậy

 \(2\cdot\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+....+\dfrac{1}{240}\right)\)

\(=2\cdot\dfrac{1}{20}+2\cdot\dfrac{1}{30}+2\cdot\dfrac{1}{42}+....+2\cdot\dfrac{1}{240}\)

\(=\dfrac{2}{2\cdot10}+\dfrac{2}{2\cdot15}+\dfrac{2}{2\cdot21}+....+\dfrac{2}{2\cdot120}\)

\(=\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+.....+\dfrac{1}{120}\)