mỗi p/số của A đều bé hơn 1/1.2+1/2.3+1/3.4+......+1/49.50

A<1-1/2+1/2-1/3+1/3-1/4+..........+1/49-1/50(tách ra thành hiệu)

A<1-1/50

mà 1/50>0=>1-1/50<1<2

A<1-1/50<1<2

A<2

chúc học tốt

\(A<\frac{1}{1\cdot2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49\cdot50}\)
          \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
          \(=1-\frac{1}{50}<1<2\)

A = 1/2.2 + 1/3.3 +......+ 1/50.50

A < 1/1.2 + 1/2.3 +......+ 1/49.50

A < 1 - 1/2 + 1/2 - 1/3 +.....+ 1/49 - 1/50

A < 1 - 1/50

A < 49/50 < 3/4 

=> A < 3/4 (đpcm)

Hình như bạn Killua giải sai thì phải.. 49/50 > 3/4 chứ

Theo mình thì bài này nên giữ nguyên phân số 1/2^2( vì nó bằng 1/4)

Xét : B = 1/3^2 + 1/4^2 +...+ 1/50^2

       => B < 1/2.3 + 1/3.4 +...+ 1/49.50

       => B<  1/2-1/3+1/3-1/4+...+1/49-1/50

       => B < 1/2-1/50 < 1/2

Suy ra A < 1/2^2 + 1/2 = 3/4 

Vậy A< 3/4

Đặt \(T=3\cdot5\cdot7\cdot.....\cdot49\)

\(\Rightarrow A\cdot T=\frac{T}{2}+\frac{T}{3}+\frac{T}{4}+....+\frac{T}{50}\)

\(2^4\cdot B\cdot T=\frac{2^4T}{2}+\frac{2^4T}{3}+\frac{2^4T}{4}+....+\frac{2^4T}{50}\left(1\right)\)

Tất cả các số hạng của (1) đều là stn ngoại trừ \(\frac{2^4T}{5}\)

\(\Rightarrow VP\notinℕ\Rightarrow VT\notinℕ\)

Mà \(2^4\inℕ\Rightarrow T\inℕ\)

\(\Rightarrow A\notinℕ\left(đpcm\right)\)

Ta có:\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4\)

\(\Rightarrow2+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4\Rightarrow\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}=2\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)

\(\Rightarrow\frac{a}{abc}+\frac{b}{abc}+\frac{c}{abc}=1\Rightarrow\frac{a+b+c}{abc}=1\Rightarrow a+b+c=abc\)

\(\Rightarrowđpcm\)

Ta có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{2}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}\)

\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)

\(\Rightarrow2^2=2+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)

\(\Leftrightarrow2=.2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)

\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1\)

\(\Leftrightarrow\frac{a}{abc}+\frac{a}{abc}+\frac{b}{abc}=\frac{abc}{abc}\)

\(\Leftrightarrow a+b+c=abc\)

\(\RightarrowĐPCM\)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)

=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1\)

=> \(\frac{a+b+c}{abc}=1\)

=> a+b+c=abc

Ta có:

\(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge a+b+c+\frac{9}{a+b+c}\)

Đặt: \(a+b+c=t\le\frac{3}{2}\Leftrightarrow2t\le3\)

Ta có: Cần cm: \(t+\frac{9}{t}\ge\frac{15}{2}\Leftrightarrow\frac{t^2+9}{t}\ge\frac{15}{2}\Leftrightarrow2t^2+18-15t\ge0\)

\(\Leftrightarrow\left(2t^2-3t\right)+\left(18-12t\right)\ge0\Leftrightarrow t\left(2t-3\right)-6\left(2t-3\right)\ge0\Leftrightarrow\left(t-6\right)\left(2t-3\right)\ge0\)(đúng với \(t\le\frac{3}{2}\))

Dấu "=" khi \(a=b=c=\frac{1}{2}\)

\(\frac{1}{2^2}< \frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2}-\frac{1}{3}\)
\(.......\)
\(\frac{1}{50^2}< \frac{1}{49}-\frac{1}{50}\)

\(\Rightarrow A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A< \frac{1}{1}-\frac{1}{50}=\frac{49}{50}\)
Mà \(\frac{49}{50}< 2\)
\(\Rightarrow A< 2\)
 

a<2 ai k cho mik, mik se k lại hứa thế lun nói là làm

ta có:1/1^2=1/1

1/2^2=1/2*2<1/1*2=1/1-1/2

1/3^2=1/3*3<1/2*3=1/2-1/3

1/4^2=1/4*4<1/3*4

...

1/50^2=1/50*50<1/49*50=1/49-1/50

=>A=1/1-1/50+1

A=99/50<100/50=2

=>A<2

vậy A<2

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

...................\(\frac{1}{50^2}< \frac{1}{49.50}\)

\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)

\(\Rightarrow A< 1-\frac{1}{50}< \frac{49}{50}< 1< 2\)

1/2^2<1/1*2;1/3^2<1/2*3;1/4^2<1/3*4;1/50^2<1/49*50

ta có:

   =>    1/1^2+1/2*3+1/3*4+...+1/49*50

  <=>   1/1-1/2+1/2-1/3+1/3-1/4+...+1/49-1/50

  <=>   1-1/50 < 2

    =>   A < 2

A=\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)

  =\(1+\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{50.50}< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

                                                                                         \(< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

                                                                                          \(< 1+1-\frac{1}{50}=\frac{99}{50}< 2\)

                     => \(A< 2\)

\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)

\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...\frac{1}{49.50}\)

\(A< 1+\frac{49}{50}\)

\(A< 1\frac{49}{50}\)

Mà \(\frac{49}{50}< 2\)nên A<2

mình làm đúng rồi

ti ck cho mình đi