1) 2300 : 50 = ( 2300 . 2 ) : ( 50 . 2 ) = 4600 : 100 = 46

2) 112 : 2 = ( 112 . 5 ) : ( 2 . 5 ) = 560 : 10 = 56

3) 2300 : 4 = ( 2300 . 25 ) : ( 4 . 25 ) = 57500 : 100 = 575

4) 3500 : 25 = ( 3500 . 4 ) : ( 4 . 25 ) = 14000 : 100 = 140

5) 1200 : 75 = ( 1200 . 4 ) : ( 4 . 75 ) = 4800 : 300 = 6

6) 2100 : 75 = ( 2100 . 4 ) : ( 4 . 75 ) = 8400 : 300 = 28

7) 320 : 8 = ( 320 . 5 ) : ( 8 . 5 ) = 1600 : 40 = 40

1) 2300 : 50 = (2300.2) : (50.2) = 4600 : 100 = 46

2) 112: 2 = ( 112.5) : (2.5) = 560 : 10 = 56

3) 2300 : 4 = ( 2300.25) : (4.25) = 57500 : 100 = 575

4)3500 : 25 = (3500.4) : (25.4) = 14000 : 100 = 140

5)1200 : 75 = ( 1200 : 3) : (75:3) = 400 : 25 = (400.4) : (25.4) = 1600 : 100 = 16

...

1) 2300 : 50 = ( 2300 . 2 ) : ( 50 . 2 ) = 4600 : 100 = 46

2) 112 : 2 = ( 112 . 5 ) : ( 2 . 5 ) = 560 : 10 = 56

3) 2300 : 4 = ( 2300 . 25 ) : ( 4 . 25 ) = 57500 : 100 = 575

4) 3500 : 25 = ( 3500 . 4 ) : ( 4 . 25 ) = 14000 : 100 = 140

5) 1200 : 75 = ( 1200 . 4 ) : ( 4 . 75 ) = 4800 : 300 = 6

6) 2100 : 75 = ( 2100 . 4 ) : ( 4 . 75 ) = 8400 : 300 = 28

7) 320 : 8 = ( 320 . 5 ) : ( 8 . 5 ) = 1600 : 40 = 40

Đáp án A

Lời giải:
$(2300-22):1+1=2279$

Tổng $A$ là:
$4+\frac{(2300+22).2279}{2}=2645923$. Số này lẻ nên không thể là lũy thừa cơ số 2. 

THI TỰ LÀM

Khiêm tốn, thật thà, dũng cảm :D

cho hỏi đi mừ

thi với thằng em đúng là thất bại khi nó là con gái

\(A=\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{100\cdot104}\)

\(A=\frac{7-4}{4\cdot7}+\frac{11-7}{7\cdot11}+\frac{15-11}{11\cdot15}+...+\frac{104-100}{100\cdot104}\)

\(A=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{100}-\frac{1}{104}\)

\(A=\frac{1}{4}-\frac{1}{104}\)

\(A=\frac{25}{104}\)

\(B=\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\)

\(B\cdot2=\left(\frac{1}{25\cdot27}+\frac{1}{27\cdot29}+\frac{1}{29\cdot31}+...+\frac{1}{73\cdot75}\right)\cdot2\)

\(B\cdot2=\frac{2}{25\cdot27}+\frac{2}{27\cdot29}+\frac{2}{29\cdot31}+...+\frac{2}{73\cdot75}\)

\(B\cdot2=\frac{27-25}{25\cdot27}+\frac{29-27}{27\cdot29}+\frac{31-29}{29\cdot31}+...+\frac{75-73}{73\cdot75}\)

\(B\cdot2=\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\)

\(B\cdot2=\frac{1}{25}-\frac{1}{75}\)

\(B\cdot2=\frac{2}{75}\)

\(B=\frac{2}{75}\frac{\cdot}{\cdot}2\)

\(B=\frac{1}{75}\)

\(C=\frac{6}{15\cdot18}+\frac{6}{18\cdot21}+\frac{6}{21\cdot24}+...+\frac{6}{87\cdot90}\)

\(\frac{C}{2}=\frac{3}{15\cdot18}+\frac{3}{18\cdot21}+\frac{3}{21\cdot24}+...+\frac{3}{87\cdot90}\)

\(\frac{C}{2}=\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+...+\frac{1}{87}-\frac{1}{90}\)

\(\frac{C}{2}=\frac{1}{15}-\frac{1}{90}\)

\(\frac{C}{2}=\frac{1}{18}\)

\(C=\frac{1}{18}\cdot2\)

\(C=\frac{1}{9}\)

a,19.21=(20-1)(20+1)=20²-1²=400-1=399

29.31=(30-1)(30+1)=30²-1²=900-1=899

39.41=(40-1)(40+1)=40²-1²=1600-1=1599

\(M=\dfrac{6}{10.13}+\dfrac{6}{13.16}+\dfrac{6}{16.19}+\dfrac{6}{19.21}\)

\(\dfrac{1}{2}M=\dfrac{3}{10.13}+\dfrac{3}{13.16}+\dfrac{3}{16.19}+\dfrac{3}{19.21}\)

\(\dfrac{1}{6}M=\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{21}\)

\(\dfrac{1}{6}M=\dfrac{1}{10}-\dfrac{1}{21}\)

\(M=\dfrac{11}{210}:\dfrac{1}{6}=\dfrac{11}{35}\)

\(N=\dfrac{1}{20}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{30}\)

\(=\dfrac{1}{20}-\dfrac{1}{30}\)

\(=\dfrac{1}{60}\)

\(\dfrac{M}{N}=\dfrac{11}{35}:\dfrac{1}{60}=\dfrac{132}{7}\)= \(\dfrac{132}{25}\)

a, \(\frac{1}{25.27}+\frac{1}{27.29}+...+\frac{1}{73.75}\)

\(=\frac{1}{2}\left(\frac{2}{25.27}+\frac{2}{27.29}+...+\frac{2}{73.75}\right)\)

\(=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+...+\frac{1}{73}-\frac{1}{75}\right)\)

\(=\frac{1}{2}\left(\frac{1}{25}-\frac{1}{75}\right)\)

\(=\frac{1}{2}\left(\frac{2}{75}\right)\)

\(=\frac{1}{75}\)

b, \(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)

\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)

\(=2\left(\frac{1004}{2010}\right)\)

\(=2\left(\frac{502}{1005}\right)\)

\(=\frac{1004}{1005}\)

Tk hộ =v

\(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}=\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)=\frac{1}{2}.\frac{2}{75}=\frac{1}{75}\)

\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)=2.\left(\frac{1}{2}-\frac{1}{2010}\right)=2.\frac{502}{1005}=\frac{1004}{1005}\)

\(\frac{1}{25.27}+\frac{1}{27.29}+\frac{1}{29.31}+...+\frac{1}{73.75}\)

\(=\frac{1}{2}.\left(\frac{2}{25.27}+\frac{2}{27.29}+\frac{2}{29.31}+...+\frac{2}{73.75}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{25}-\frac{1}{75}\right)\)

\(=\frac{1}{2}.\left(\frac{3}{75}-\frac{1}{75}\right)\)

\(=\frac{1}{2}.\frac{2}{75}\)

\(=\frac{1}{75}\)

Câu dưới đặt 2 ra ngoài rồi làm bình thường.