Question

1) Tính nhanh:

P=\(1\frac{1}{3}\cdot1\frac{1}{8}\cdot1\frac{1}{15}\cdot1\frac{1}{24}\cdot1\frac{1}{35}\cdot1\frac{1}{48}\cdot1\frac{1}{63}\cdot1\frac{1}{80}\)

2) So sánh:

A=\(\frac{100^{10}+1}{100^{10}-1}\) và B=\(\frac{100^{10}-1}{100^{10}-3}\)

3) So sánh A và B biết:

A=\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)

B=\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)

Answer 1

#It's the moment when you're in good mood, you accidentally click back =.=

1) Calculate

\(P=1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}....1\frac{1}{63}.1\frac{1}{80}\)

\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{64}{63}.\frac{81}{80}\)

\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}....\frac{8.8}{7.9}.\frac{9.9}{8.10}\)

\(=\frac{2.9}{10}=\frac{9}{5}\)

Answer 2

ta có: 100¹⁰ + 1 > 100¹⁰ - 1

⇒ A = \(\frac{100^{10}+1}{100^{10}-1}< \frac{100^{10}+1-2}{100^{10}-1-2}=\frac{100^{10}-1}{100^{10}-3}=B\)

vậy A < B

Answer 3

3)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1-\frac{49}{50}\)

\(=\frac{49}{50}\)

⇒ A < 1 (1)

\(B=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)

\(\Rightarrow B>\frac{1}{10}+\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)=\frac{1}{10}+\frac{90}{100}=1\)

⇒ B > 1 (2)

từ (1) và (2) ⇒ A<1<B

vậy A < B