a) A = 3.4 + 4.5 + 5.6 + ...+ 49.50

=> 3A = 3.4.3+4.5.3+ 5.6.3+...+49.60.3

3A = 3.4.(5-2) +4.5.(6-3) + 5.6.(7-4) + ...+ 49.60.(61-48)

3A = 3.4.5 - 2.3.4 + 4.5.6 -3.4.5 + 5.6.7-4.5.6 + 49.60.61 - 48.49.60

3A = -2.3.4 + 49.60.61

\(A=\frac{-2.3.4+49.60.61}{3}=59772\)

b) B = 1.3 + 3.5 + 5.7 + ...+ 51.53

=> 6B = 1.3.6 + 3.5.6 + 5.7.6 + ...+ 51.53.6

6B = 1.3.(5+1) + 3.5.(7-1) + 5.7.(9-3) +...+ 51.53.(55-49)

6B = 1.3.5 + 1.3 + 3.5.6 - 1.3.5 + 5.7.9 - 3.5.7 + ...+ 51.53.55 - 49.51.53

6B = 1.3 + 51.53.55

\(B=\frac{1.3+51.53.55}{6}=24778\)

cau c mk ko bk

d) D = 1 + 3 + 9 + 27 + 81 + 243 + 729 + 2187 + 6561

D = 3⁰+3¹+3²+3³+3⁴+3⁵+3⁶+3⁷+3⁸

=> 3D = 3¹+3²+3³+...+3⁸+3⁹

=> 3D - D = 3⁹-3⁰

2D = 3⁹-1

\(D=\frac{3^9-1}{2}=9841\)

làm dài lắm,nếu muốn thì k minh còn ko thì thôi

a,0,36.350+1,2.20.3+9.4.4,5

=13.3.35+12.2.3+9.2.3.3

=3.(13.35+12.2+.9.2.3)

=3.(455+24+54)

=3.533

=1599

b,2015.2016-5/2015.2015+2010

=4062240-5+2010

=4064245

c,2/1.3+2/3.5+2/5.7+...+2/71.73

=1-1/3+1/3-1/5+1/5-1/7+...+1/71-1/73

=1-1/73

=72/73

d,(1+1/2).(1+1/3)+...+(1+1/2018)

=3/2.4/3.5/4+...+2019/2018

=2019/2

e,E=1/4.5+1/5.6+1/6.7+...+1/80.81(làm tương tự với phần d nên mình làm ngắn

     =1/4-1/81

     =77/324

f,F=3/2.3+3/3.4+...+3/99.100

=3.(1/2.3+1/3.4+...+1/99.100)(làm tương tự với d

=3.(1/2-1/100)

=3.49/100

=147/100

gG=5/1.4+5/4.7+...+5/61.64

3G=5.(3/1.4+3./4.7+...+3/61.64)

     =5.(1-1/64)

     =5.63/64

     =315/64

ok nha bạn,mình giữ đúng lời hứa.

           \(4.B=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{93.97}\) 

            \(4.B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{97}\)

            \(4.B=1-\frac{1}{97}\)

             \(4.B=\frac{96}{97}\)

                 \(B=\frac{96}{97}:4\)

                 \(B=\frac{24}{97}\)

b)

\(\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}\)

\(=\frac{32}{99}\)

c)

\(\frac{7}{3.4}+\frac{7}{4.5}+.....+\frac{7}{60.61}\)

\(=7\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{60}-\frac{1}{61}\right)\)

\(=7\left(\frac{1}{3}-\frac{1}{61}\right)\)

\(=\frac{406}{183}\)

d)

\(\frac{6}{2.4}+\frac{6}{4.6}+....+\frac{1}{72.74}\)

\(=3\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{72}-\frac{1}{74}\right)\)

\(=3\left(\frac{1}{2}-\frac{1}{74}\right)\)

=57/37

Cô giáo à bạn bị sao thế. Ng` ta có bài kg biết thì phải hỏi , đằng này bạn lại còn đi chửi ng` khác nữa. Đây chắc chắn là giáo viên giả mạo rồi.

cô giáo bạn đúng là hk phải gv thiệt. mất dạy quá!!!

(a+\(\dfrac{1}{1.3}\))+(a+\(\dfrac{1}{3.5}\))+(a+\(\dfrac{1}{5.7}\))+..+(a+\(\dfrac{1}{23.25}\))=11.a+(\(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)+\(\dfrac{1}{243}\))

(a+a+..+a)+(\(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{23.25}\)) = 11.a+ \(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)+\(\dfrac{1}{243}\))

Đặt A =(a+a+..+a) + \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{23.25}\)

Xét dãy số 1; 3; 5;...;25 Dãy số trên là dãy số cách đều với khoảng cách là: 3-1 = 2

Dãy số trên có số số hạng là: (25 - 1): 2 + 1  = 13

Vậy A = a\(\times\)13 + \(\dfrac{1}{1.3}\)+\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+...+\(\dfrac{1}{23.25}\)

A = a\(\times\)13 + \(\dfrac{1}{2}\) \(\times\)(\(\dfrac{2}{1.3}\)+\(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+...+\(\dfrac{2}{23.25}\))

A = a \(\times\) 13 + \(\dfrac{1}{2}\times\)( \(\dfrac{1}{1}-\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)- \(\dfrac{1}{7}\)+...+\(\dfrac{1}{23}\) - \(\dfrac{1}{25}\))

A = a\(\times\)13 + \(\dfrac{1}{2}\) \(\times\) \(\dfrac{24}{25}\)

A = a\(\times\)13 + \(\dfrac{12}{25}\) (1)

Đặt B =    \(\dfrac{1}{3}\) + \(\dfrac{1}{9}\)+ \(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)+\(\dfrac{1}{243}\)

B\(\times\)3 =1 + \(\dfrac{1}{3}\)+\(\dfrac{1}{9}\)+\(\dfrac{1}{27}\)+\(\dfrac{1}{81}\)

B\(\times\)3 - B = 1 - \(\dfrac{1}{243}\) = \(\dfrac{242}{243}\)

2B = \(\dfrac{242}{243}\)

B = \(\dfrac{242}{243}\): 2

B = \(\dfrac{121}{243}\)

11a + B = 11a + \(\dfrac{121}{243}\) (2)

Từ (1) và(2) ta có:

a\(\times\)13  + \(\dfrac{12}{25}\) = 11\(\times\) a + \(\dfrac{121}{143}\)

a \(\times\) 13 + \(\dfrac{12}{25}\) - 11 \(\times\)a = \(\dfrac{121}{143}\) 

\(a\times\)(13 - 11) + \(\dfrac{12}{25}\) = \(\dfrac{121}{143}\)

a \(\times\) 2 + \(\dfrac{12}{25}\) = \(\dfrac{121}{243}\)

a \(\times\) 2 = \(\dfrac{121}{243}\) - \(\dfrac{12}{25}\)

a \(\times\) 2 = \(\dfrac{109}{6075}\)

a = \(\dfrac{109}{6075}\): 2

a = \(\dfrac{109}{12150}\)

Bài 5:

a) Ta có: \(A=1\cdot2+2\cdot3+3\cdot4+...+9\cdot10\)

\(\Leftrightarrow3\cdot A=3\cdot\left(1\cdot2+2\cdot3+3\cdot4+...+9\cdot10\right)\)

\(\Leftrightarrow3A=1\cdot2\cdot\left(3-0\right)+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+9\cdot10\cdot\left(11-8\right)\)

\(\Leftrightarrow3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+8\cdot9\cdot10-8\cdot9\cdot10+9\cdot10\cdot11\)

\(\Leftrightarrow3\cdot A=9\cdot10\cdot11=90\cdot11=990\)

hay A=330

Vậy: A=330