CMR :
\(\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+5+7}+...+\frac{1}{1+3+5+...+2017}< \frac{3}{4}\)
Giúp mk nha
CMR :
\(\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+5+7}+...+\frac{1}{1+3+5+...+2017}< \frac{3}{4}\)
Giúp mk với
\(S=\frac{1}{1+3}+\frac{1}{1+3+5}+...+\frac{1}{1+3+5+7+...+2017}\)
\(S=\frac{1}{\left[\left(1+3\right):2\right]^2}+\frac{1}{\left[\left(1+5\right):2\right]^2}+...+\frac{1}{\left[\left(2017+1\right):2\right]^2}\)
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1009^2}\)
\(S< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1007.1008}\)
\(S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1008}-\frac{1}{1009}\)
\(S< \)
Còn đâu làm nốt , tao đi ngủ đây
Ai giúp mk nha:
\(ChoA=\frac{1}{1+3}+\frac{1}{1+3+5}+...+\frac{1}{1+3+5+...+2013}\)
CMR Nó <\(\frac{3}{4}\)
Cho \(A=\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+5+7}+...+\frac{1}{1+3+5+7+...+2013}\)
Cm nó <\(\frac{3}{4}\)Nha mn giúp mk nha còn có mỗi bài này nữa
tính hợp lý
a) \(\frac{-1}{2}+\frac{-1}{9}-\frac{-3}{5}+\frac{1}{2006}-\left(\frac{-2}{7}\right)-\frac{7}{18}+\frac{4}{35}\)
b) \(\frac{1}{3}-\frac{3}{4}+\frac{3}{5}+\frac{1}{2007}-\frac{1}{36}+\frac{1}{15}-\frac{2}{9}\)
giúp mk nha
a) \(\frac{-1}{2}+\frac{-1}{9}-\frac{-3}{5}+\frac{1}{2006}-\frac{-2}{7}-\frac{7}{18}+\frac{4}{35}\)
\(=\left(\frac{-1}{2}-\frac{1}{9}-\frac{7}{18}\right)+\left(\frac{3}{5}+\frac{4}{35}\right)+\frac{1}{2006}\)
\(=\left(\frac{-9}{18}-\frac{2}{18}-\frac{7}{18}\right)+\left(\frac{21}{35}+\frac{4}{35}\right)+\frac{1}{2006}\)
\(=\left(\frac{-9-2-7}{18}\right)+\left(\frac{21+4}{35}\right)+\frac{1}{2006}\)
\(=\left(\frac{-18}{18}\right)+\left(\frac{25}{35}\right)+\frac{1}{2006}\)
\(=\left(-1\right)+\frac{5}{7}+\frac{1}{2006}\)\(=\frac{-4005}{14042}\)
b) \(\frac{1}{3}-\frac{3}{4}+\frac{3}{5}+\frac{1}{2007}-\frac{1}{36}+\frac{1}{15}-\frac{2}{9}\)
\(=\left(\frac{1}{3}+\frac{1}{2007}-\frac{2}{9}\right)-\left(\frac{3}{4}+\frac{1}{36}\right)+\left(\frac{3}{5}+\frac{1}{15}\right)\)
\(=\left(\frac{669}{2007}+\frac{1}{2007}-\frac{446}{2007}\right)-\left(\frac{27}{36}+\frac{1}{36}\right)+\left(\frac{9}{15}+\frac{1}{15}\right)\)
\(=\frac{224}{2007}-\frac{28}{36}+\frac{10}{15}\)
\(=\frac{224}{2007}-\frac{1561}{2007}+\frac{1338}{2007}\)\(=\frac{1}{2007}\)
Chứng tỏ rằng:
\(\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+5+7}+...+\frac{1}{1+3+5+7+...+2017}< \frac{3}{4}\)
tính
1/\(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{9}}{\left(\frac{1}{4}+\frac{1}{7}-\frac{-3}{35}\right).\left(-1\frac{1}{3}\right)}\)
2/\(\frac{0,6-\frac{1}{3}+\frac{3}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)
các bn lm giúp mk vs!!!!
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ai lm nhanh nhất đúng đủ trình bày khoa học mk tick cho !!!!!!!1
\(A=\frac{3}{7}-\frac{3}{17}+\frac{3}{37}:\frac{5}{7}-\frac{5}{17}+\frac{5}{37}\) + \(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}:\frac{7}{5}-\frac{7}{4}-\frac{7}{3}-\frac{7}{2}\)
Tính hợp lí nha:
(Dấu : là phần nha)
Làm ơn giúp mình nha
Cho A= \(\frac{1}{1+3}+\frac{1}{1+3+5}+\frac{1}{1+3+5+7}+...+\frac{1}{1+3+5+7+...+2017}\)
Chứng minh A<\(\frac{3}{4}\)
A=1/(1+3)+1/(1+3+5)+1/(1+3+5+7)+...+1/(1+3+5+7+...+2017)
A=1/2^2+1/3^2+1/4^2+...+1/1009^2
2A=2/2^2+2/3^2+2/4^2+...+2/1009^2
Ta co :(x-1)(x+1)=(x-1)x+x-1=x^2-x+x-1=x^2-1<x^2
suy ra 2A<2/(1*3)+2/(3*5)+2/(5*7)+...+2/(1008*1010)
suy ra 2A <1-1/3+1/3-1/5+1/5-1/7+...+1/1008-1/1010
suy ra 2A<1-1/1010
suy ra 2A<2009/2010<1<3/2
suy ra 2A <3/2
suy ra A <3/4 (dpcm)
nho k cho minh voi nha
A=1/(1+3)+1(1+3+5)+1/(1+3+5+7)+....+1/(1+3+5+7+...+2017)
A=1/4+1/9+1/16+....+1/1018081
A=1/2^2+1/3^2+1/4^2+...+1/1009^2
Ta có : 1/3^2=1/3x3<1/2x3
1/4^2=1/4x4<1/3x4
......
1/1009^2<1/1008x1009
Suy ra 1/2^2+1/3^2+1/4^2+.....+1/1009^2<1/2^2+1/2x3+1/3x4+.....+1/1008x1009
Suy ra A< 1/2^2+1/2-1/3+1/3-1/4+.....+1/1008-1/1009
=> A<1/2^2+1/2+1/3-1/3+......+1/1008-1/1008-1/1009
=> A<1/2^2+( 1/2-1/1009)
=> A< 3023/4036
Mà +) 3023<3/4
+) A<3023/4026
Suy ra A<3/4
=> A<1008/1009
Ta có 1008/1009+
so sánh 2 số A và B nếu
\(A=-\frac{1}{2018}-\frac{3}{2017^2}-\frac{5}{2017^3}-\frac{7}{2017^4};B=\frac{-1}{2018}-\frac{7}{2017^2}-\frac{5}{2017^3}-\frac{3}{2017^4}\)