A= 1,01-1,02-1,03-...-1,98

Mà từ 0, 01 đến 1, 98 có 98 số số hạng

A= 1,01-1,02-1,03-...-1,98

=> (1,01-1,02)-1,03-...-(1,97-1,98)

Khi nhóm 2 số lại với nhau thành 49 cặp số số hạng

=> (-0, 01) -.......- (-0, 01)

mà -0, 01 có 49 số

Ta lấy (-0, 01) .49= (-0,49)

Vậy A= (-0, 49)

\(a,A=\dfrac{101}{100}+\dfrac{102}{100}+\dfrac{103}{100}+...+\dfrac{199}{100}\)

\(A=\dfrac{101+102+103+...+109}{100}\)

Xét tử số : \(101+102+103+...+199\)

Có : \(\left(199-101\right):1+1=99\) (số hạng)

\(\Rightarrow\) Tử số bằng \(:\left(199+101\right).99:2=14850\)

\(\Rightarrow A=\dfrac{14850}{100}=\dfrac{297}{2}\)

\(b,B=\dfrac{10002}{10000}+\dfrac{10004}{10000}+\dfrac{10006}{10000}+...+\dfrac{12014}{10000}\)

\(B=\dfrac{10002+10004+10006+...+12014}{10000}\)

\(B=\dfrac{10002+10004+10006+...+12014}{10000}\)

Xét tử số : \(10002+10004+10006+...+12014\)

Có : \(\left(12014-10002\right):2+1=1007\) (số hạng)

\(\Rightarrow\) Tử số bằng : \(\left(12014+10002\right).1007:2=11085056\)

\(\Rightarrow B=\dfrac{11085056}{10000}\)

Bạn tự làm câu C nha

\(D=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{2014.2015}\)

\(\Rightarrow D=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{2014}-\dfrac{1}{2015}\)

\(\Rightarrow D=\dfrac{1}{5}-\dfrac{1}{2015}=\dfrac{402}{2015}\)

\(E=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{2014.2017}\)

\(\Rightarrow3E=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{2014.2017}\)

\(\Rightarrow3E=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{2014}-\dfrac{1}{2017}\)

\(\Rightarrow3E=1-\dfrac{1}{2017}=\dfrac{2016}{2017}\)

\(\Rightarrow E=\dfrac{2016}{2017}:3=\dfrac{672}{2017}\)

D = \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\) +...+ \(\dfrac{1}{2014.2015}\)
D = \(\dfrac{1}{5}\) - \(\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)+...+ \(\dfrac{1}{2014}-\dfrac{1}{2015}\)
D = \(\left(\dfrac{1}{5}-\dfrac{1}{2015}\right)\)
D = \(\dfrac{403}{2015}-\dfrac{1}{2015}\)
D = \(\dfrac{402}{2015}\)

Đáp án :C=15,5

kq= 15,5.

\(a,\Leftrightarrow4\left(x-2\right)=3\left(x+3\right)\left(x\ne-3\right)\\ \Leftrightarrow4x-8=3x+9\\ \Leftrightarrow x=17\left(tm\right)\\ b,\Leftrightarrow\dfrac{-1,04}{x}=\dfrac{18,72}{16,38}\\ \Leftrightarrow x=\dfrac{-1,04\cdot16,38}{18,72}=-0,91\)

a) ĐKXĐ: \(x\ne-3\)

\(\Leftrightarrow4x-8=3x+9\\ \Leftrightarrow x=17\)

b) \(\Leftrightarrow-1,04:x=\dfrac{8}{7}\\ \Leftrightarrow x=-\dfrac{91}{100}\)

150 + 1,03 : [ 10,3 . ( x - 1 ) ] = 160

         1,03 : [ 10,3 . ( x - 1 ) ] = 160 - 150

         1,03 : [ 10,3 . ( x - 1 ) ] = 10 

                  [ 10,3 . ( x - 1 ) ] = 1,03 : 10

                  [ 10,3 . ( x - 1 ) ] = 0,103

                              ( x - 1 )  = 0,103 : 10,3

                              ( x - 1 )   =  0,01

                                x           = 0,01 + 1

                                 x          = 1,01

b) Từ đề bài ,ta có: \(\hept{\begin{cases}\frac{-\left(x+3\right)}{27}=\frac{-121}{33}\left(1\right)\\\frac{11}{1-2y}=\frac{-121}{33}\left(2\right)\end{cases}}\)

Giải (1):

\(\frac{-\left(x+3\right)}{27}=\frac{-121}{33}=>-\left(x+3\right)=-121.27:33=-99=>-x-3=-99=>-x=-96=>x=96\)

Giải (2) :

\(\frac{11}{1-2y}=\frac{-121}{33}=>1-2y=11.33:\left(-121\right)=-3=>2y=4=>y=2\)

Vậy x=96;y=2