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\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}< 1\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{49}-\frac{1}{50}< 1\)
\(S=1-\frac{1}{50}< 1\)
\(S=\frac{49}{50}< 1\left(đpcm\right)\)
Chứng tỏ rằng;\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+......+\(\frac{1}{49.50}\) < 1
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1-\frac{1}{50}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{49}-\frac{1}{49}\right)\)
\(=\left(1-\frac{1}{50}\right)+0+0+...+0=1-\frac{1}{50}=\frac{50}{50}-\frac{1}{50}=\frac{49}{50}\)
Vì \(\frac{49}{50}<1\) nên \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}<1\)
Vậy thỏa mãn đề bài
Ta có:
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
Mà \(\frac{49}{50}< 1\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}< 1\left(đpcm\right)\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{1}-\frac{1}{50}\)
\(=\frac{49}{50}\)
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
Ta có: \(\frac{1}{1.2}=\frac{3}{1.2.3}\) ;\(\frac{1}{1.2+2.3}=\frac{3}{2.3.4}\); \(\frac{1}{2.3+3.4}=\frac{3}{3.4.5}\); ......;\(\frac{1}{1.2+2.3+3.4+...+n\left(n+1\right)}=\frac{3}{n\left(n+1\right)\left(n+2\right)}\)
=> \(S=\frac{3}{1.2.3}+\frac{3}{2.3.4}+\frac{3}{3.4.5}+...+\frac{3}{n\left(n+1\right)\left(n+2\right)}\)
=> \(\frac{2S}{3}=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
Ta lại có: \(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3}\); \(\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4}\); \(\frac{2}{3.4.5}=\frac{1}{3.4}-\frac{1}{4.5}\);....;\(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
=> \(\frac{2S}{3}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
=> \(\frac{2S}{3}=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)=> \(S=\frac{3}{4}-\frac{3}{2\left(n+1\right)\left(n+2\right)}< \frac{3}{4}\)
=> \(S< \frac{3}{4}\)
\(=1-\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{99}+\frac{1}{99}\right)-\frac{1}{100}\)
\(=1+0+0+...+0-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}< 11\)
Vậy : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}< 11\)
=1/2-1/3+1/3-1/4+...+1/99-1/100
=1/2-1/100
=50/100-1/100
=49/100<1
=> dãy trên < 1 đđcm
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 < \(\frac{1}{2}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(\Rightarrow A< 1\)
\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=\frac{1}{2}-\frac{1}{100}\)
\(\Rightarrow B< \frac{1}{2}\)
A= 1/1-1/2+1/2-1/3+1/4-1/5+...+1/101-1/102
A=1-1/102=102/102-1/102=101/102
ý b thì chờ mình tí tìm cách lập luận đã nhé
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}+\frac{1}{101.102}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{102}\)
\(A=1-\frac{1}{102}\)
\(A=\frac{101}{102}\)
`Answer:`
Bài 1:
a. \(\frac{1}{2}-\left(\frac{2}{3}x-\frac{1}{3}\right)=\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{2}-\frac{2}{3}x+\frac{1}{3}=\frac{2}{3}\)
\(\Leftrightarrow\frac{5}{6}-\frac{2}{3}x=\frac{2}{3}\)
\(\Leftrightarrow-\frac{2}{3}=\frac{2}{3}-\frac{5}{6}\)
\(\Leftrightarrow-\frac{2}{3}x=-\frac{1}{6}\)
\(\Leftrightarrow x=-\frac{1}{6}:-\frac{2}{3}\)
\(\Leftrightarrow x=\frac{1}{4}\)
b. \(\frac{3}{x+5}=15\%\left(ĐKXĐ:x\ne-5\right)\)
\(\Leftrightarrow\frac{3}{x+5}=\frac{3}{20}\)
\(\Leftrightarrow\frac{60}{20\left(x+5\right)}=\frac{3\left(x+5\right)}{20\left(x+5\right)}\)
\(\Leftrightarrow60x=3x+15\)
\(\Leftrightarrow-3x=-45\)
\(\Leftrightarrow x=15\)
Bài 2:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{4}+\frac{1}{4}\right)+...+\left(-\frac{1}{49}+\frac{1}{49}\right)-\frac{1}{50}\)
\(=1+0+0+0+...+0-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{50}{50}-\frac{1}{50}\)
\(=\frac{49}{50}< 1\)
Vậy A < 1
KO BIẾT