A = \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+ ... + \(\frac{1}{308}\)+ \(\frac{1}{309}\)

B = \(\frac{308}{1}\)+ \(\frac{307}{2}\)+ \(\frac{306}{3}\)+\(\frac{3}{306}\) + \(\frac{2}{307}\)+ \(\frac{1}{308}\)

=> B = \(\frac{309-1}{1}\)+ \(\frac{309-3}{3}\)+... + ( 309 ... )

=> B = 309 + 309 . ( \(\frac{1}{2}\) + \(\frac{1}{3}\)+... + \(\frac{1}{306}\)+ \(\frac{1}{307}\)+ \(\frac{1}{308}\)+ \(\frac{1}{309}\)- \(\frac{1}{1}\)+ \(\frac{2}{2}\)+ ... + \(\frac{308}{308}\)+ \(\frac{309}{309}\)

=> B = 309 . ( \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ ... + \(\frac{1}{306}\)+ \(\frac{1}{307}\)+ \(\frac{1}{308}\)+ \(\frac{1}{309}\))

=> \(\frac{A}{B}\)= \(\frac{1}{309}\)

Lâu rồi bạn còn cần lời giải ko mình giải cho

Ta có :

\(B=\frac{308}{1}+\frac{307}{2}+\frac{306}{3}+...+\frac{3}{306}+\frac{2}{307}+\frac{1}{308}\)

\(B=\left(\frac{307}{2}+1\right)+\left(\frac{306}{3}+1\right)+...+\left(\frac{3}{306}+1\right)+\left(\frac{2}{307}+1\right)+\left(\frac{1}{308}+1\right)+1\)

\(B=\frac{309}{2}+\frac{309}{3}+...+\frac{309}{306}+\frac{309}{307}+\frac{309}{308}+\frac{309}{309}\)

\(B=309.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{306}+\frac{1}{307}+\frac{1}{308}+\frac{1}{309}\right)\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{308}+\frac{1}{309}}{309.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{308}+\frac{1}{309}\right)}\)

\(\frac{A}{B}=\frac{1}{309}\)

\(B=308/1+307/2+306/3+...+1/308 \)

\(B=308+307/2+306/3+...+1/308\) chia số 308 thành 308 số 1

B=307/2+1+306/3+1+...+1/308+1+1

B=309/2+309/3+309/4+...+309/308+309/309

B=309(1/2+1/3+1/4+...+1/309)=309A

Suy ra A/B=1/309

=(1/2+1/31/4...1/307/1/3081/309)/(309-1/1+309-2/2+...+309-307/307+309-308/308)

=(1/21/31/4...1/3071/3081/309)/(309/1-1+309/2-1+...+309/307-1+309/308-1)

=(........................................)/(309/309309/2309/3...309/307+309/308)

=(........................................)/[309x(1/309+1/308+...+1/41/31/2)]

Thấy tử và mẫu giống nhau thì ta rút:

=1/309

a. Ta có :

B = 308/1 + 307/2 +306/3+....+1/308

B = (1+1+....+1) ₊ 307/2 + ....+ 1/308

B = (1 + 307/2) + (1+306/3) + ...+ (1+ 1/308) + 1

B = 309/2 + 309/3 + ....+ 309/308 + 309/309

B = 309.(1/2 + 1/3 + ....+1/309)

Vậy A/B: 1/2 + 1/3 + ... + 1/309 / 308/1 + 307/2 +....+ 2/307+1/308

       A/B = 1/2 + 1/3 +... + 1/309 /  309.(1/2 + 1/3 + ....+1/309)

       A/B = 1/309

b.7/10.11 + 7/11.12 + .... +7 /69.70

= 7. (1/10.11+1/11.12 + ...+ 1/69.70)

= 7.(1/10-1/11+1/11-1/12+....+1/69-1/70)

= 7.(1/10 - 1/70)

= 7. 3/35

= 3/5

ĐặtA= \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{308}+\frac{1}{309}\)

     \(\frac{1}{A}=1\div\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{308}+\frac{1}{309}\right)\)

=> \(\frac{1}{A}=2+3+4+...+308+309\)

=>Ta có: Số các số hạng là:(309-2)/1+1=308(số hạng) 

              Tổng của\(\frac{1}{A}\)là:\(\frac{\left(309+2\right).308}{2}\)=47894

=> \(\frac{1}{A}=47894\)

=>\(A=\frac{1}{47894}\)

Chúc bạn học tốt!

2a)

Gọi số cần tìm là abc.

Để abc = a.

Theo đề bài, ta có: a chia 25 dư 5 => a - 20 chia hết cho 25

a chia 28 dư 8 => a - 20 chia hết cho 28

a chia 35 dư 15 => a - 20 chia hết cho 35

Vậy a - 20 \(\in\)BC (25, 28, 35)

25 = 5²

28 = 2² . 7

35 = 5 . 7

BCNN (25, 28, 35) = 5² . 2² . 7 = 700

a - 20 \(\in\)BC (25, 28, 35)

mà BC (25, 28, 35) = B (700)

nên a - 20 \(\in\) B (700) = {0 ; 700 ; 1400 ; 2800 ; ...}

Vậy a \(\in\){680 ; 1380 ; 2780 ; ...}

mà a là số có ba chữ số.

=> abc = 680.

Vậy số tự nhiên cần tìm là 680.

bài này dễ nhưng mất tg lắm k mình trước mình làm sau cho