\(S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{100}\)

\(\Rightarrow2S=2+1+\frac{1}{2}+\frac{1}{2^2}...+\frac{1}{99}\)

\(2S-S=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)

\(\Leftrightarrow2S-S=S=2-\frac{1}{2^{100}}=\frac{2^{101}}{2^{100}}-\frac{1}{2^{100}}=\frac{2^{101}-1}{2^{100}}\)

\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)

\(>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

\(\Rightarrow S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{2}{5}\)

nhưng tại sao lại >1/2*3+1/3*4+1/4*5+...+1/9*10

bạn giúp mk nha

\(-4\frac{1}{2}\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le\frac{-2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)

<=>  \(-1,5\le x\le\frac{11}{18}\)

đến đây tự làm

mk k biết điều kiện của x  nên giúp đến đó

= 1/2 . 2/3 .... 2014/2015 . 2015/2016

= 1/2016

1/2016

A = ( 1 - 1/2 ) ( 1 - 1/3 ) ... ( 1 - 1/2015 ) ( 1 - 1/2016 )

A = 1/2 . 2/3 ... 2014/2015 . 2015/2016

A = 1 . 2 . 3 ... 2014 . 2015/2 . 3 ... 2014 . 2015 . 2016

A = 1/2016

Nếu đúng thì k cho mình nha

\((6-4\frac{1}{2}):0,03+(3\frac{1}{20}-2,65):4+\frac{2}{5}\)

\((6-4,5):0,03+(3,05-2,65):4+0,4\)

\(=1,5:0,03+0,4:4+0,4\)

\(=50+(0,4\cdot2):4\)

\(=50+0,2\)

\(=50,2\)

Chúc bạn học tốt

\(\frac{1}{3}\left(x-1\right)+\frac{2}{5}\left(x+1\right)=0\)

\(\frac{1}{3}x-\frac{1}{3}+\frac{2}{5}x+\frac{2}{5}=0\)

\(\frac{11}{15}x+\left(\frac{2}{5}-\frac{1}{3}\right)=0\)

\(\frac{11}{15}x+\frac{1}{15}=0\)

\(\frac{1}{15}\left(11x+1\right)=0\)

\(11x+1=0\)

\(\Rightarrow x=-\frac{1}{11}\)

\(\frac{a}{1}=\frac{b}{2}=\frac{c}{3}=\frac{5a}{5}=\frac{2b}{4}=\frac{8c}{24}=\frac{5a+2b+8c}{5+4+24}\)(*)

\(\frac{a}{1}=\frac{b}{2}=\frac{c}{3}=\frac{-3a}{-3}=\frac{-4b}{-8}=\frac{6c}{18}=\frac{-3a-4b+6c}{-3-8+18}\)(**)

Lấy (*) chia cho (**) được kết quả: A=\(\frac{7}{33}\)

Đăng từ bài thôi bạn à!

a) Áp dụng công thức: \(\frac{1}{a-1}-\frac{1}{a}=\frac{1}{\left(a-1\right)a}>\frac{1}{a.a}=\frac{1}{a^2}\)

Ta có: \(\frac{1}{2^2}< \frac{1}{1}-\frac{1}{2}\)

\(\frac{1}{3^2}< \frac{1}{2}-\frac{1}{3}\)

\(\frac{1}{4^2}< \frac{1}{3}-\frac{1}{4}\)

..............................

\(\frac{1}{n^2}< \frac{1}{n-1}-\frac{1}{n}\)

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\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1-\frac{1}{n}=\frac{1}{n+1}< 1\)

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1\) (đpcm)