viết ..... thế thì ai mà biết

12.A= 1.5.12+5.9(13-1)+9.13(17-5)+13.17(21-9)+.....+97.101(105 - 93)

12.A = 1.5.12 + 5.9.13 -1.5.9 + 9.13.17- 5.9.13 +.....+ 97.101.105 -93.97.101

12.A = 1.5.12 -1.5.9 + 97.101.105

A =  (1.5.12 -1.5.9 + 97.101.105):12 = 85725

\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{97.101}\)

\(=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{101}\right)\)

\(=\frac{1}{4}.\left(1-\frac{1}{101}\right)\)

\(=\frac{1}{4}.\frac{100}{101}\)

\(=\frac{25}{101}\)

TA CÓ : a,bc . a+b+c = 10

=> a+b+c = 10 : a,bc 

=> a+b+c = 10 -2 (Do a,bc = a+b+c )

=> a+b+c = 8 = a,bc

Mà a,bc . a+b+c = 10

=> a =1 , b=2 , c= 5

Thử thay vào nhé

1/1.5 + 1/5.9 + 1/9.13 + ... + 1/97.101

= 1/4.(4/1.5 + 4/5.9 + 4/9.13 + ... + 4/97.101)

= 1/4.(1 - 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/97 - 1/101)

= 1/4.(1 - 1/101)

= 1/4.100/101

= 25/101

\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+........+\frac{1}{97.101}\)

\(=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+........+\frac{1}{97}-\frac{1}{101}\right)\)

\(=\frac{1}{4}\left(1-\frac{1}{101}\right)\)

\(=\frac{1}{4}.\frac{100}{101}=\frac{25}{101}\)

\(A=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{97.101}\)

\(4A=\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{97.101}\)

\(4A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{101}\)

\(4A=1-\frac{1}{101}=\frac{100}{101}\)

\(A=\frac{100}{101}:4=\frac{25}{101}\)

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nếu bn cho mk là dúng thì mk trả lời cho

AI

K

CHO

MINH

VOI

CAM

ON