Bài 1: 

\(\Leftrightarrow-\dfrac{5}{7}:x=-\dfrac{7}{18}-\dfrac{1}{6}=\dfrac{-7}{18}-\dfrac{3}{18}=\dfrac{-10}{18}=\dfrac{-5}{9}\)

=>x=5/9:5/7=7/9

Bài 2:

a: \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{1000}{999}=\dfrac{1000}{2}=500\)

b: \(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-999}{1000}\)

\(=-\dfrac{1}{1000}\)

Câu c:

C = 3/2^2.8/3^2.15/4^2...99/10^2

C = \(\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times\ldots\times\) \(\frac{9.11}{10.10}\)

C = \(\frac{1.2.3\ldots9}{2.3\ldots10}\) x \(\frac{3.4.\ldots11}{2.3.\ldots10}\)

C = \(\frac{1}{10}\) x \(\frac{11}{2}\)

C = 11/20

Bài 1:

1/6 + -5/7 : x = - 7/18

5/7: x = 1/6 + 7/18

5/7: x = 3/18 + 7/18

5/7: x = 5/9

x = 5/7 : 5/9

x = 5/7 x 9/5

x = 9/7

Vậy x = 9/7

Bài 2:

Câu a:

(1/2 + 1).(1/3 + 1).(1/4 + 1)...(1/999 + 1)

= (1/2 + 2/2).(1/3 + 3/3)...(1/999 + 999/999)

= 3/2.4/3....1000/999

= 1000/2

= 500

biết làm bài 1 thôi

\(\left(\frac{1}{2}+1\right)\times\left(\frac{1}{3}+1\right)\times\cdot\cdot\cdot\times\left(\frac{1}{999}+1\right)\)

= \(\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times\cdot\cdot\cdot\times\frac{1000}{999}\)

lượt bỏ đi còn :

\(\frac{1000}{2}=500\)

Bài 2:

B = (1/2 - 1)(1/3 -1).(1/4 -1)...(1/1000 - 1)

B = (1/2 - 2/2).(1/3 - 3/3)...(1/1000 - 1000/1000)

B = (-1/2).(-2/3)...(-999/1000)

Xét dãy số: 1; 2 ;3;...999

Dãy số trên có 999 số hạng vậy B là tích của 999 số âm

B = - 1/1000

a,\(\frac{31}{1000}\)

b,\(\frac{8}{108}\)

c,0

a,\(\frac{49}{97}\)

b,\(\frac{-1}{4751}\)

a)

\(=\frac{3}{2}.\frac{4}{3}......\frac{100}{99}=\frac{100}{2}=50\)

b)

\(=\frac{\left(-1\right)}{2}.\frac{\left(-2\right)}{3}.....\frac{\left(-99\right)}{100}=\frac{-1}{100}\)

\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)

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

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

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

\(T=2.\frac{502}{1005}=\frac{1004}{1005}\)

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

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009+2011}\)

\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2009+2011}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)

\(A=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)

\(A=\frac{1}{2}.\frac{2010}{2011}\)

\(\Rightarrow A=\frac{1005}{2011}\)

\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)

\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)

1 A=\(\frac{31}{60}\)

2B=c,\(\frac{26\cdot32^{7}}{21}\approx4.25406\cdot10^{10}\)

3 C<\(\frac{1}{21}\)

4 D<\(\frac{11}{19}\)

\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{999}-1\right)=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-998}{999}=\frac{1}{999}\)

-------

\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{99}{10^2}=\frac{3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{9.11}{10.10}=\frac{\left(2.3.4...9\right).\left(3.4.5...11\right)}{\left(2.3.4...10\right).\left(2.3.4...10\right)}=\frac{1.11}{10.2}=\frac{11}{20}\)

E cứ xem bài của Trà My nhé. Cậu ấy làm gần đúng rồi đó