mk ko piết tính nhưng pn nên tìm ra quy tắc pn sẽ lm dc

Câu 1,2,3 Ez quá rồi :3

Câu 4:

Tổng quát:

\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v

Câu 5 ko khác câu 4 lắm :v

Câu 5: 

Tổng quát:

\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v

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

-------

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

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

Câu a :

A = 2\(\frac12\) : (- \(\frac12\))^2 - \(\frac{1}{-3}\).(\(\frac{1}{-2}\) - \(\) \(\frac43\) : \(-\frac89\))

A = \(\frac52\) : \(\frac14\) + \(\frac13\).(-\(\frac12\) + \(\frac43\times\frac98\))

A = \(\frac52\times4\) + \(\frac13\).(- \(\frac12+\frac32\))

A = \(10\) + 1/3

A = 30/3 + 1/3

A = 31/3

Câu b:

B = (3\(\frac{10}{99}\) + 4\(\frac{11}{99}\) - \(\frac{58}{229}\)).(\(\frac12\) - \(\frac43\) - \(\frac16\))

B = (7\(\frac{21}{99}\) - \(\frac{58}{229}\))(3/6 - 8/6 - 1/6)

B = (7\(\frac{7}{33}\) - \(\frac{58}{229}\)).(-1)

B = - (\(\frac{238}{33}-\frac{58}{229}\))

B = - (\(54502\)/7557 - 1914/7557)

B = - 52588/7557

A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

A < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

A < \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

A < 1 - \(\frac{1.}{100}\)

A < \(\frac{99}{100}< \frac{199}{100}\)

=> A < \(\frac{199}{100}\)

b,

S = \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{99}{10^2}\)

S = \(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{9.11}{10.10}\)

S = \(\frac{1.3.2.4.3.5.4.6.5.7...9.11}{2.2.3.3.4.4...10.10}\)

S = \(\frac{1.2.3^2.4^2.5^2...9^2.10.11}{2^2.3^3.4^2...10^2}\)

S = \(\frac{1.11}{2.10}\)

S = \(\frac{11}{20}\)

2, \(\frac{10}{1.2.3}+\frac{10}{2.3.4}+\frac{10}{3.4.5}+....+\frac{10}{100.101.102}\)

  \(=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{102-100}{100.101.102}\)

  \(=\frac{10}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{100.101}-\frac{1}{101.102}\right)\)

  \(=\frac{10}{2}.\left(\frac{1}{1.2}-\frac{1}{101.102}\right)\)

  \(=\frac{10}{2}.\frac{2575}{5151}\)

  \(=2,499514657\)

= 2,499514657 bạn nhé