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M\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{49}-\frac{1}{50}\)
\(1-\frac{1}{50}=\frac{49}{50}\)
vì \(\frac{49}{50}
A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
=\(1-\frac{1}{50}\)
Vì \(1-\frac{1}{50}< 1\)nên A < 1
B = \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
=\(\frac{1}{2}-\frac{1}{100}\)
Vì \(\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)nên B < \(\frac{1}{2}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(\Rightarrow A< 1\)
\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=\frac{1}{2}-\frac{1}{100}\)
\(\Rightarrow B< \frac{1}{2}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}< 1\)
\(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{49\times50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
So sánh \(\frac{49}{50}< 1\)nên \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{49\times50}< 1\)
Ta có : 1/1.2 + 1/2.3 + ... + 1/49.50
= 1-1/2+1/2-1/3 +...+1/49-1/50
= 1- 1/50
= 49/50 > 45/50 = 9/10(đpcm)
\(\Rightarrow M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow M=\frac{1}{1}-\frac{1}{50}\)
\(\Rightarrow M=\frac{49}{50}\)
\(\Rightarrow\frac{49}{50}<1\)
\(\Rightarrow M<1\)
M = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)
M = \(1-\frac{1}{50}\)
M = \(\frac{49}{50}<1\)
=> M < 1
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
Mà \(\frac{49}{50}\)lại nhỏ hơn 1 \(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}< 1\)
\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}< 1\)
1/1.2+1/2.3+1/3.4+.....+1/49.50
=1-1/2+1/2-1/3+1/3-1/4+....+1/49-1/50
=1-1/50
=49/50
\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(M=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(M=\frac{1}{1}-\frac{1}{50}=\frac{49}{50}\)
Vì \(\frac{49}{50}\)< 1 nên M < 1.
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#Sunrise
\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(M=1-\frac{1}{50}\)
\(M=\frac{50}{50}-\frac{1}{50}\)
\(M=\frac{49}{50}\)\(< \frac{50}{50}\)
\(M< 1\)
Chúc bạn học tốt nha !!!