Bài 2:

1: \(x+\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\cdots+\frac{1}{31\cdot33}=3\)

=>\(x+\frac12\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{31\cdot33}\right)=3\)

=>\(x+\frac12\left(1-\frac13+\frac13-\frac15+\cdots+\frac{1}{31}-\frac{1}{33}\right)=3\)

=>\(x+\frac12\left(1-\frac{1}{33}\right)=3\)

=>\(x+\frac12\cdot\frac{32}{33}=3\)

=>\(x+\frac{16}{33}=3\)

=>\(x=3-\frac{16}{33}=\frac{99}{33}-\frac{16}{33}=\frac{83}{33}\)

2: \(x-\frac{3}{1\cdot5}-\frac{3}{5\cdot9}-\cdots-\frac{3}{61\cdot65}=2\)

=>\(x-\frac34\left(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\cdots+\frac{4}{61\cdot65}\right)=2\)

=>\(x-\frac34\left(1-\frac15+\frac15-\frac19+\cdots+\frac{1}{61}-\frac{1}{65}\right)=2\)

=>\(x-\frac34\left(1-\frac{1}{65}\right)=2\)

=>\(x-\frac34\cdot\frac{64}{65}=2\)

=>\(x-\frac{48}{65}=2\)

=>\(x=2+\frac{48}{65}=\frac{130}{65}+\frac{48}{65}=\frac{178}{65}\)

Bài 1:

2: \(B=\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\cdots+\frac{1}{89\cdot90}\)

\(=\frac16-\frac17+\frac17-\frac18+\cdots+\frac{1}{89}-\frac{1}{90}\)

\(=\frac16-\frac{1}{90}=\frac{14}{90}=\frac{7}{45}\)

3: \(C=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\cdots+\frac{1}{60\cdot62}\)

\(=\frac12\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\cdots+\frac{2}{60\cdot62}\right)\)

\(=\frac12\left(\frac12-\frac14+\frac14-\frac16+\cdots+\frac{1}{60}-\frac{1}{62}\right)\)

\(=\frac12\left(\frac12-\frac{1}{62}\right)=\frac12\cdot\frac{30}{62}=\frac12\cdot\frac{15}{31}=\frac{15}{62}\)

4: \(D=\frac{2}{2\cdot5}+\frac{2}{5\cdot8}+\cdots+\frac{2}{92\cdot95}\)

\(=\frac23\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\cdots+\frac{3}{92\cdot95}\right)\)

\(=\frac23\left(\frac12-\frac15+\frac15-\frac18+\cdots+\frac{1}{92}-\frac{1}{95}\right)\)

\(=\frac23\left(\frac12-\frac{1}{95}\right)=\frac23\cdot\frac{93}{190}=\frac{1}{95}\cdot31=\frac{31}{95}\)

5:Sửa đề: \(E=\frac{6}{1\cdot3}+\frac{6}{3\cdot5}+\cdots+\frac{61}{63\cdot65}\)

\(=3\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{63\cdot65}\right)\)

\(=3\left(1-\frac13+\frac13-\frac15+\cdots+\frac{1}{63}-\frac{1}{65}\right)\)

\(=3\left(1-\frac{1}{65}\right)=3\cdot\frac{64}{65}=\frac{192}{65}\)

Đặt \(A=0,01+0,02+0,03+...+0,1\)

\(100A=100.\left(0,01+0,02+0,03+...+0,1\right)\)

\(100A=1+2+3+...+10\)

\(100A=\frac{\left(10+1\right).10}{2}\)

\(100A=\frac{11.10}{2}\)

\(100A=55\)

\(A=\frac{55}{100}\)

\(A=\frac{11}{20}\)

\(A=0,55\)

Vậy \(0,01+0,02+0,03+...+0,1=0,55\) 

Ta có:

1 + 3 + 5 + 7 +...+2000001

Các số 1,2,3,5,7,.....,2000001 lập thành dãy số tự nhiên cách đều có khoảng cách là 2 đơn vị

                                       Số số hạng của dãy là:

                                                    (   2000001 -  1 )  : 2 + 1 = 1000001 ( số hạng )                                                                

                                        Tổng của dãy trên là :                             

                                                     (  2000001 + 1 ) x 1000001 :2 = 1000002000001

                                                                                                  Đ/s: 1000002000001    
 

117/38    |          2,3      |         4,75       |        2,5 |

1.  3.078947368421053 giờ
2.  2.3giờ
3.  4.75 giờ
4.  2.5 giờ

Tổng 2 vận tốc:

56+48=104(km/h)

Hai ô tô sẽ gặp nhau sau thời gian là:

260:104=2,5(giờ)=2 giờ 30 phút

Chọn A

8D 5D 6B

A

9 km 26 m = 9,026 km

892mm = 0,892 mm

4m 783 mm = 47,83 dm

106 m = 1,06hm

8 dm 9 mm = 0,809m

=55,19/24+24+19,22

=2,2+24+19,22

=26,2+19,22

=45,42

sai đề rồi bạn nhé

\(6\frac{2}{15}:3\frac{5}{7}=\frac{92}{15}:\frac{26}{7}=\frac{92}{15}\times\frac{7}{26}=\frac{322}{195}\)

\(6\frac{2}{15}\)\(:3\frac{5}{7}\)

\(=\frac{92}{15}\)\(:\frac{26}{7}\)

\(=\frac{92}{15}\)\(\times\frac{26}{7}\)

\(=\frac{322}{195}\)