Bài 11: Tính chất cơ bản của phép nhân phân số

x có nghĩa là nhân nha

Câu đầu sai đề nhé! Phải là 2007 chứ ko phải 20007!

\(A=\dfrac{1}{2}\cdot\dfrac{1}{7}+\dfrac{1}{7}\cdot\dfrac{1}{12}+...+\dfrac{1}{2002}\cdot\dfrac{1}{2007}\\ =\dfrac{1}{2\cdot7}+\dfrac{1}{7\cdot12}+...+\dfrac{1}{2002\cdot2007}\\ =\dfrac{1}{5}\left(\dfrac{5}{2\cdot7}+\dfrac{5}{7\cdot12}+...+\dfrac{5}{2002+2007}\right)\\ =\dfrac{1}{5}\cdot\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+...+\dfrac{1}{2002}-\dfrac{1}{2007}\right)\\ =\dfrac{1}{5}\left(\dfrac{1}{2}-\dfrac{1}{2007}\right)\\ =\dfrac{1}{5}\cdot\dfrac{2005}{4014}\\ =\dfrac{401}{4014}\)

\(B=\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)...\left(1+\dfrac{1}{2007}\right)\\B=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\cdot\cdot\dfrac{2008}{2007}\\ B=\dfrac{3\cdot4\cdot...\cdot2008}{2\cdot3\cdot...\cdot2007}\\ B=\dfrac{2008}{2}\\ B=1004 \)

\(C=\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2008}\right)\\ =\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2007}{2008}\\ =\dfrac{1\cdot2\cdot...\cdot2007}{2\cdot3\cdot...\cdot2008}\\ =\dfrac{1}{2008}\)

Để A đạt giá trị nguyên thì :

\(n+1⋮n-2\)

Mà \(n-2⋮n-2\)

\(\Leftrightarrow3⋮n-2\)

\(\Leftrightarrow n-2\inƯ\left(3\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}n-2=1\\n-2=-1\\n-2=-3\\n-2=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}n=3\\n=1\\n=-1\\n=5\end{matrix}\right.\)

Vậy ..

Tính tổng sau bằng phương pháp hợp lí nhất:

A=\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{2450}\);

B=\(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{1443}\)

Gọi thời gian đi là: a (a > 0)
Thời gian về là: b (b > 0)
a + b = 3,5 (h) ⇒ a = 3,5 - b
Do độ dài quãng đường BC không đổi nên ta được:
30a = 40b
Hay 30(3,5 - b) = 40b
⇒ 105 - 30b = 40b
⇒ 105 = 70b
⇒ b = 1,5
Quãng đường BC là: 40.1,5 = 60 (km)

Bài này áp dụng quy tắc chuyển vế là ra

\(\left(42-98\right)-42-12\)

\(=\left(42-42\right)-\left(98+12\right)\)

\(=0-110\)

\(=-110\)

(42-98)-42-12

= (42-42)-98+12

=0-110

= -110

Chúc bạn học giỏi !!!!

(42-98)-42-12

= (42-42)-98+12

=0-110

= -110

Chúc bn hc tốt

A)17/5.-31/125.1/2.10/17.-1/2 mũ 3

=(17/5.10/17).(1/2.-1/6).-31/125

=2.(-1/12).(-31/125)

=-1/6.(-31/125)

=31/125

B)(11/4.-5/9.4/9.11/4).8/33

=(11/4.(-5/9.4/9)).8/33

=(11/4.(-1)).8/33

=(-11/4).8/33

=-2/3

C)(135/21 +45/123 - 55/434).(-5/12 + 1/4 + 1/6

=(135/21 +45/123 - 55/434).(5/12+3/12+2/12)

=(135/21 +45/123 - 55/434).0

= 0

D) 132/111. (1 /2 + 1/3 + 7/6)

=12.(3/6+2/6+7/6)

= 12.7/3

= 28

E)1/5.1/6 + 1/6.1/7 + 1/7.1/8 + 1/8 . 1/9 + 1/9.9/10

=1/5.6+1/6.7+1/7.8+1/8.9+(1/9.9/10)

=1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/10

=1/5-1/10

= 1/10

Mình ko biết làm câu F thông cảm nhé

Sai thì bình luận cho mình biết với nhé

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\\ =\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\\ =\frac{1}{2}-\frac{1}{100}=\frac{50}{100}-\frac{1}{100}\\ =\frac{49}{100}\)

Thay b vào, ta có:\(B=\dfrac{3}{4}.\dfrac{6}{19}+\dfrac{4}{3}.\dfrac{6}{19}-\dfrac{1}{2}.\dfrac{6}{19}=\dfrac{1}{2}\)

Thay c vào, ta có:\(C=\dfrac{2002}{2003}.\dfrac{3}{4}+\dfrac{2002}{2003}.\dfrac{5}{6}-\dfrac{2002}{2003}.\dfrac{19}{12}=0\)

a: \(=\dfrac{2}{3}+\dfrac{1}{3}\cdot\dfrac{-8+15}{18}:17\)

\(=\dfrac{2}{3}+\dfrac{1}{3}\cdot\dfrac{7}{18}\cdot\dfrac{1}{17}\)

\(=\dfrac{619}{918}\)

b: \(=\left(3-7+\dfrac{1}{4}\right)\cdot\left(4-\dfrac{31}{6}+\dfrac{9}{4}\right)\)

\(=\dfrac{-15}{4}\cdot\dfrac{13}{12}=\dfrac{-195}{48}=\dfrac{-65}{16}\)

c: \(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}\cdot\dfrac{12}{5}-\dfrac{1}{4}\)

\(=\dfrac{1}{2}-\dfrac{3}{2}+\dfrac{6}{5}=\dfrac{6}{5}-1=\dfrac{1}{5}\)

d: \(=\dfrac{5}{4}\cdot\dfrac{1}{4}:\left(\dfrac{21}{16}-\dfrac{3}{2}\right)\)

\(=\dfrac{5}{16}:\dfrac{-3}{16}=\dfrac{-5}{3}\)

1.\(\left(\dfrac{1}{2}+1\right)\cdot\left(\dfrac{1}{3}+1\right)\cdot\left(\dfrac{1}{4}+1\right)..\left(\dfrac{1}{999}+1\right)\)

\(=\left(\dfrac{1}{2}+\dfrac{2}{2}\right)\cdot\left(\dfrac{1}{3}+\dfrac{3}{3}\right)\cdot...\left(\dfrac{1}{999}+\dfrac{999}{999}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{1000}{999}\)\(=\dfrac{3\cdot4\cdot5\cdot...\cdot1000}{2\cdot3\cdot4\cdot...\cdot999}\)

\(=\dfrac{1000}{2}=500\).

2.

\(\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{1000}-1\right)\)

\(=\left(\dfrac{1}{2}-\dfrac{2}{2}\right)\cdot\left(\dfrac{1}{3}-\dfrac{3}{3}\right)...\left(\dfrac{1}{1000}-\dfrac{1000}{1000}\right)\)

Thôi mai mk làm tiếp nha