Chứng tỏ:
\(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}>2\)
Giải chi tiết giúp mk nka
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1/2+1/3+1/4+...+1/63>1/31+1/31+...+1/31(62 số hạng 1/31)
hay 1/2+1/3+1/4+...+1/63>62 x 1/31
nên 1/2+1/3+1/4+...+1/63>2(dpcm)
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
\(>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
\(\Rightarrow S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{2}{5}\)
Gọi UCLN(12n+1,30n+2)=d(d thuộc N*)
=>12n+1 chia hết d => 60n+5 chia hết d (1)
30n+2 chia hết d => 60n+4 chia hết d (2)
Lấy (1)-(2) : 60n+5- 60n -4=1 chia hết d => d thuộc ước của 1
=> 12n+1/30n+2 là phân số tối giản
_ giải bừa :v _
\(T=\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{14^2}\)
Ta thấy : \(\frac{1}{4^2}< \frac{1}{2.4};\frac{1}{14^2}< \frac{1}{12.14}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{14^2}< \frac{1}{2^2}+\frac{1}{2.4}+...+\frac{1}{12.14}\)
\(\Rightarrow T< \frac{1}{2^2}+\frac{1}{2}\left(\frac{2}{2.4}+...+\frac{2}{12.14}\right)\)
\(\Rightarrow T< \frac{1}{2^2}+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{14}\right)\)
\(\Rightarrow T< \frac{1}{4}+\frac{1}{2}.\frac{3}{7}\)
\(\Rightarrow T< \frac{13}{28}\)
Mà \(\frac{13}{28}< \frac{1}{2}\Rightarrow T< \frac{1}{2}\)
....
\(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\)
\(\left(x+1\right)\left(x+3\right)=\left(0,5x+2\right)\left(2x+1\right)\)
\(x^2+4x+3=x^2+4,5x+2\)
\(x^2-x^2+4x-4,5x-2+3=0\)
\(1-0,5x=0\)
\(x=2\)
1/2=1/2
1/3+1/4>1/4+1/4=1/2
1/5+…+1/8>4*1/8=1/2
1/9+…+1/16>8*1/16=1/2
1/2+1/3+1/4+…+1/16>4*1/2=2
1/2+1/3+1/4+…+1/63>1/2+1/3+1/4+…+1/16
=> 1/2+1/3+…+1/63>2
t i c k nhé !! 5756876876978080
Ta có:
\(\frac{1}{2}=\frac{1}{2}\)
\(\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)
\(\frac{1}{5}+...+\frac{1}{8}>4.\frac{1}{8}=\frac{1}{2}\)
\(\frac{1}{9}+...+\frac{1}{16}>8.\frac{1}{16}=\frac{1}{2}\)
\(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{16}>4.\frac{1}{2}=2\)
\(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}>\frac{1}{2}+\frac{1}{3}+...+\frac{1}{16}\)
\(\Rightarrow\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}>2\)