Ta có : \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};\frac{1}{4^2}<\frac{1}{3.4};...;\frac{1}{2011^2}<\frac{1}{2010.2011};\frac{1}{2012^2}<\frac{1}{2011.2012}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2010+2011}+\frac{1}{2011.2012}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}+\frac{1}{2011}-\frac{1}{2012}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}<\frac{1}{1}-\frac{1}{2012}\)

Vì \(\frac{1}{2012}>0\) => \(\frac{1}{1}-\frac{1}{2012}<1\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}<1\)

zee that tri tuệ ve toan day so; ok 10đ

Ta có

\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};.....;\frac{1}{2012^2}< \frac{1}{2011.2012}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2012^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2011.2012}\)

= 1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2011}-\frac{1}{2012}\)

=1-\(\frac{1}{2012}\)=\(\frac{2011}{2012}< 1\)

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{2012^2}< 1\)

a ) 

Theo bài ra: (a - 4) chia hết cho 5 => (a - 4) + 20 chia hết cho 5 => a + 16 chia hết cho 5

(a - 5) chia hết cho 7 => (a - 5) + 21 chia hết cho 7 => a + 16 chia hết cho 7

(a - 6) chia hết cho 11 => (a - 6) + 22 chia hết cho 11 => a + 16 chia hết cho 11 

=> a + 16 thuộc BC(5; 7; 11) 

Mà BCNN(5; 7; 11) = 385

=> a + 16 thuộc B(385) = {0; 385; 770; ...}

=> a thuộc {-16; 369; 754;...}

Vì a là số tự nhiên nhỏ nhất

=> a = 369 

b ) \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.......+\frac{1}{2011^2}+\frac{1}{2012^2}.\)

Ta có : 

\(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)

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

\(\frac{1}{2012^2}=\frac{1}{2012.2012}< \frac{1}{2011.2012}\)

Ta có :

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.......+\frac{1}{2011^2}+\frac{1}{2012^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2011.2012}\)

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.......+\frac{1}{2011^2}+\frac{1}{2012^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2011}-\frac{1}{2012}\)

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.......+\frac{1}{2011^2}+\frac{1}{2012^2}< 1-\frac{1}{2012}\)

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.......+\frac{1}{2011^2}+\frac{1}{2012^2}.< \frac{2011}{2012}\)

Mà \(\frac{2011}{2012}< 1\)

\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.......+\frac{1}{2011^2}+\frac{1}{2012^2}< 1\)

\(b)\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}\)​

\(< \)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{2010.2011}+\frac{1}{2011.2012}\)

\(< \)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2012}\)

\(< \)\(1-\frac{1}{2012}\)\(=\frac{2011}{2012}< 1\)

Vậy Biểu thức    \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2011^2}+\frac{1}{2012^2}\)\(< 1\)

\(a)\)

Theo bài ra: \(\left(a-4\right)⋮5\Rightarrow\left(a-4\right)+20⋮5\Rightarrow a+16⋮5\)

\(\left(a-5\right)⋮7\Rightarrow\left(a-5\right)+21⋮7\Rightarrow a+16⋮7\)

\(\left(a-6\right)⋮11\Rightarrow\left(a-6\right)+22⋮11\Rightarrow a+16⋮11\)

\(\Rightarrow\) \(a+16\in BC\left(5;7;11\right)\)

Mà \(BCNN\left(5;7;11\right)=385\)

\(\Rightarrow\) \(a+16\in B\left(385\right)=\left\{0;385;770;...\right\}\)

\(\Rightarrow\) \(a\in\left\{-16;369;754;...\right\}\)

Vì a là số tự nhiên nhỏ nhất \(\Rightarrow\) \(a=369\)

c)\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\)

\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{2012}}\right)\)

\(2A=2+1+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2011}}\)

\(2A-A=\left(2+1+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....\frac{1}{2^{2012}}\right)\)

\(A=2-\frac{1}{2^{2012}}\)

1/

A=1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100

A=1/1-1/100

Vì 1/100>0

-->1/1-1/100<1

-->A<1

a)\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)

\(\frac{1}{1}-\frac{1}{100}\)=\(\frac{99}{100}<1\)

Ta có:\(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2012^2}>0\)

Vì:  \(\frac{1}{2^2}<\frac{1}{1.2}\)

\(\frac{1}{3^2}>\frac{1}{2.3}\)

\(\frac{1}{4^2}>\frac{1}{3.4}\)

..........

\(\frac{1}{2012^2}>\frac{1}{2011.2012}\)

\(\Rightarrow A<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2011.2012}\)

\(\Rightarrow A<1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}\)

\(\Rightarrow A<1-\frac{1}{2012}\)

\(\Rightarrow A<1\)

Vì A>0;A<1

=>A không phải số tự nhiên

=>ĐPCM

Quy đồng A lên thì tử số chia hết cho 2011² còn mẫu số không chia hết cho 2011² vì có \(\frac{1}{2011^2}\) khi quy đồng thì tử không chia hết cho 2011²

Vậy A không phải là số tự nhiên

chọn đúng cho mk nha

2.+ \(\left(2n+1\right)^2=4n^2+4n+1>4n^2+4n\)

\(\Rightarrow2n+1>\sqrt{4n\left(n+1\right)}=2\sqrt{n\left(n+1\right)}\)

+ \(\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(2n+1\right)\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

Do đó : \(A< \frac{1}{2}\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{48}}-\frac{1}{\sqrt{49}}\right)\)

\(\Rightarrow A< \frac{1}{2}\)

1. + \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\left(n+1\right)-n}{\left(n+1\right)\sqrt{n}}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(n+1\right)\sqrt{n}}\)

\(< \frac{\left(\sqrt{n+1}-\sqrt{n}\right)\cdot2\sqrt{n+1}}{\sqrt{n}\left(n+1\right)}=2\cdot\frac{n+1-\sqrt{n\left(n+1\right)}}{\left(n+1\right)\sqrt{n}}=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

Do đó : \(A< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)

\(\Rightarrow A< 2\)

Bài 2 tạm thời chưa nghĩ ra :))

\(\frac{B}{A}=\frac{\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{1}{2012}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}\)

\(=\frac{\left(\frac{2011}{2}+1\right)+\left(\frac{2010}{3}+1\right)+...+\left(\frac{1}{2012}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}\)

\(=\frac{\frac{2013}{2}+\frac{2013}{3}+\frac{2013}{4}+....+\frac{2013}{2012}+\frac{2013}{2013}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2013}}\)

\(=\frac{2013\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2013}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}=2013\)

Ta có : 

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2011^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2010.2011}\)\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}=1-\frac{1}{2011}=\frac{2010}{2011}>\frac{2010}{2680}=\frac{3}{4}\)

Hình như có gì đó sai sai :')

A+1/4=1/2+1/3²+......+1/2011²

A+1/4<1/2+1/2*3 +1/3*4 +....1/2010*2011

A+1/4<1-1/2011<1=3/4+1/4

A<1/4 (ĐPCM)

NHẦM BẠN ƠI     A<3/4