A = 5 + 5\(^2\) + ...+ 5\(^{50}\)

A x 5 = 5\(^2\) + 5\(^3\) + ... + 5\(^{51}\)

A x 5 - A = 5\(^2\) + 5\(^3\) + ... + 5\(^{51}\) - 5 - 5\(^2\) -..-5\(^{50}\)

A x (5 - 1) = (5\(^2\) - 5\(^2\))+..+(5\(^{50}-5^{50}\)) + (5\(^{51}\)- 5)

A x 4 = 0 + 0 + .. + 0 + 5\(^{51}\) - 5

A x 4 = 5\(^{51}\) - 5

A = (5\(^{51}\) - 5)/4

A = 5 + 5\(^2\) + ...+ 5\(^{50}\)

A = 5(1 + 5 + ... + 5\(^{49}\)) ⋮ 5 (đpcm)

A = 5 + 5\(^2\) + ...+ 5\(^{50}\)

Xét dãy số: 1; 2;...; 50

Dãy số trên là dãy số cách đều với khoảng cách là: 2 - 1 = 1

Số số hạng của dãy số trên là:

(50 - 1) : 1 + 1 = 50(số hạng)

Vì 50 : 2 = 25

Nên nhóm hai số hạng liên tiếp của A vào nhau ta được:

A = (5 + 5\(^2\)) + .. + (5\(^{49}\) + 5\(^{50}\))

A = 5(1 + 5) + ... + 5\(^{49}\).(1 + 5)

A = 5.6 + ... + 5\(^{49}\).6

A = 6.(5 + ... + 5\(^{49}\)) ⋮ 6 (đpcm)

a: Ta có: \(A=5+5^2+5^3+\cdots+5^{49}+5^{50}\)

=>\(5A=5^2+5^3+\cdots+5^{51}\)

=>\(5A-A=5^2+5^3+\cdots+5^{51}-5-5^2-\cdots-5^{50}\)

=>\(4A=5^{51}-5\)

=>\(A=\frac{5^{51}-5}{4}\)

b: Ta có: \(A=5+5^2+5^3+\cdots+5^{49}+5^{50}\)

\(=5\left(1+5+5^2+\cdots+5^{48}+5^{49}\right)\) ⋮5

c: ta có: \(A=5+5^2+5^3+\cdots+5^{49}+5^{50}\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\cdots+\left(5^{49}+5^{50}\right)\)

\(=5\left(1+5\right)+5^3\left(1+5\right)+\cdots+5^{49}\left(1+5\right)\)

\(=6\left(5+5^3+\cdots+5^{49}\right)\) ⋮6

A = 5 + 5 ^ 2 + 5 ^ 3 + ... + 5 ^ 50

5 A = 5 ^ 2 + 5 ^ 3 + 5 ^ 4 + ... + 5 ^ 51

5 A - A = ( 5 ^ 2 + 5 ^ 3 + 5 ^ 4 + ... + 5 ^ 51 )

            -  ( 5 + 5 ^ 2 + 5 ^ 3 + ... + 5 ^ 50 )

4 A      = 5 ^ 51 - 5

A         = \(\frac{5^{51}-5}{4}\)

A=5^1+5^21+5^3+...+5^50

5^1A=5(5^1+5^2+5^3+..+5^50)

5A=5^2+5^3+..+5^50+5^51

5A-A=(5^2+5^3+..+5^50+5^51)-(5^1+5^2+5^3+..+5^50)

4A=5^51-5^1

A=(5^51-5^1):4

=(5^51-5):25

\(A=2^0+2^1+2^2\)\(+2^3+...+\)\(2^{50}\)

\(2A=2+2^2+2^3+...+2^{51}\)

\(2A-A=A=2^{51}-2^0\)

\(B=5+5^2+5^3+...+5^{99}+5^{100}\)

\(5B=5^2+5^3+5^4+...+5^{100}+5^{101}\)

\(5B-B=4B=5^{101}-5\)

\(B=\frac{5^{101}-5}{4}\)

\(C=3-3^2+3^3-3^4+...+\)\(3^{2007}-3^{2008}+3^{2009}-3^{2010}\)

\(3C=3^2-3^3+3^4-3^5+...-3^{2008}+3^{2009}-3^{2010}+3^{2011}\)

\(3C+C=4C=3^{2011}+3\)

\(C=\frac{3^{2011}+3}{4}\)

\(S_{100}=5+5\times9+5\times9^2+5\times9^3+...+5\times9^{99}\)

\(S_{100}=5\times\left(1+9+9^2+9^3+...+9^{99}\right)\)

\(9S_{100}=5\times\left(9+9^2+9^3+...+9^{99}+9^{100}\right)\)

\(9S_{100}-S_{100}=8S_{100}=5\times\left(9^{100}-1\right)\)

\(S_{100}=\frac{5\times\left(9^{100}-1\right)}{8}\)

A=20+21+22+23+...++23+...+250250

2�=2+22+23+...+2512A=2+22+23+...+251

2�−�=�=251−202A−A=A=251−20

�=5+52+53+...+599+5100B=5+52+53+...+599+5100

5�=52+53+54+...+5100+51015B=52+53+54+...+5100+5101

<...

\(A=1+5+5^2+..+5^{49}+5^{50}\)

\(5A=5+5^2+5^3+...+5^{50}+5^{51}\)

\(5A-A=\left(5+5^2+5^3+...+5^{51}\right)-\left(1+5+5^2+...+5^{50}\right)\)

\(4A=\left(5-5\right)+\left(5^2-5^2\right)+...+\left(5^{50}+5^{50}\right)+5^{51}-1\)

\(4A=0+0+...+0+5^{51}-1\)

\(4A=5^{51}-1\)

\(A=\frac{5^{51}-1}{4}\)

Câu a:

\(\frac{2^{12}.3^5-4^6.9^2}{12^6+8^4.3^5}\)

= \(\frac{2^{12}.3^5-2^{12.3^6}}{2^{12}.3^6+2^{12}.3^5}\)

= \(\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}\)

= \(\frac{2}{3.4}\)

= \(\frac16\)

Câu b:

\(\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

= \(\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)

= \(\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+8\right)}\)

= \(\frac{5.\left(-6\right)}{9}\)

=- 10/3

Ai nhanh mik k thề đó

 Thu gọn : 5 + 5^2 + 5^3 + ... + 5^49 + 5^50

đặt tên biểu thức trên là A

ta có :

\(A=5+5^2+5^3+...+5^{49}+5^{50}\)

\(5A=5.\left(5+5^2+5^3+...+5^{49}+5^{50}\right)\)

\(5A=5^2+5^3+5^4+...+5^{50}+5^{51}\)

\(5A-A=\left(5^2+5^3+5^4+...+5^{50}+5^{51}\right)-\left(5+5^2+5^3+...+5^{49}+5^{50}\right)\)

\(4A=5^{51}-5\)

\(A=\left(5^{51}-5\right):4\)