2:

a: =>101y+5050=5555

=>101y=505

=>y=5

b: =>3^y+1=3^5

=>y+1=5

=>y=4

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{26.27}\)

=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{26}-\dfrac{1}{27}\)

=\(1-\dfrac{1}{27}\)

=\(\dfrac{26}{27}\)

Bài làm:

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)

\(\Leftrightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(\Leftrightarrow A=1-\frac{1}{10}\)

\(\Leftrightarrow A=\frac{9}{10}\)

Vậy \(A=\frac{9}{10}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{1}-\frac{1}{10}\)

\(=\frac{10}{10}-\frac{1}{10}\)

\(=\frac{9}{10}\)

P/s: E nên lưu ý các dạng bài này nhé! Đây thường là câu cuối trong đề thi cuối kì đấy!

\(A=1.2+2.3+3.4+...+n.\left(n+1\right)\)

\(\Rightarrow3A=1.2.3+2.3.4+3.4.3+...+3n.\left(n+1\right)\)

\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n.\left(n+1\right).\left[\left(n+2\right)-\left(n-1\right)\right]\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+n.\left(n+1\right).\left(n+2\right)-\left(n-1\right)n.\left(n+1\right)\)

\(3A=n.\left(n+1\right).\left(n+2\right)\)

\(\Rightarrow A=\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)

Vậy \(A=\frac{n.\left(n+1\right).\left(n+2\right)}{3}.\)

Chúc em học tốt!

3A=1.2.3 + 2.3.3 + 3.4.3 +... + n.(n+1).3

=1.2.(3-0) + 2.3.(4-1) + ... + n.(n+1).[(n+2)-(n-1)]

=[1.2.3+ 2.3.4 + ...+ (n-1).n.(n+1)+ n.(n+1)(n+2)] - [0.1.2+ 1.2.3 +...+(n-1).n.(n+1)]

=n.(n+1).(n+2)

=>S=[n.(n+1).(n+2)] /3

Mình gõ câu a bị lỗi nha , thực chất câu a là

a) Tìm các số tự nhiên x, y biết : 2xy + x + 2y = 13

a)Bạn làm nha vì bài này dễ rồi

b)+)Ta có:A=1.2+2.3+3.4+..................+99.100

=>3A=1.2.3+2.3.3+3.4.3+.................+99.100.3

=>3A=1.2.3+2.3.(4-1)+3.4.(5-2)+................+99.100.(101-98)

=>3A=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...................-98.99.100+99.100.101

=>3A=99.100.101

=>A=\(\frac{99.100.101}{3}=333300\)

+)Ta lại có:B=1²+2²+3²+..................+99²

=>B=1.1+2.2+3.3+............+99.99

=>B=1.(2-1)+2.(3-1)+3.(4-1)+..........+99.(100-1)

=>B=1.2-1+2.3-2+3.4-3+........................+99.100-99

=>B=(1.2+2.3+3.4+............+99.100)-(1+2+3+..............+99)

Đặt N=1.2+2.3+3.4+....................+99.100

=>3N=1.2.3+2.3.3+3.4.3+.................+99.100.3

=>3N=1.2.3+2.3.(4-1)+3.4.(5-2)+................+99.100.(101-98)

=>3N=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...................-98.99.100+99.100.101

=>3N=99.100.101

=>N=\(\frac{99.100.101}{3}=333300\)

Đặt M=1+2+3+..............+99(có 99 số hạng)

=>M=\(\frac{\left(1+99\right).99}{2}=4950\)

+)Ta thấy A-B=333300-(333300-4950)

=>A-B=333300-333300+4950

=>A-B=4950\(⋮\)50

Vậy A-B\(⋮\)50

Chúc bn học tốt

\(x=1.2+2.3+3.4+...+99.100\)

\(x=1.\left(1+1\right)+2.\left(2+1\right)+3.\left(3+1\right)+...+99.\left(99+1\right)\)

\(x=1^2+1.1+2^2+2.1+3^2+3.1+...+99^2+99.1\)

\(x=y+\left(1+2+3+...+99\right)\)

\(x=y+\frac{99.\left(99+1\right)}{2}=y+4950\)

\(x-y=4950\)