Ta đặt:A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...\frac{1}{n^2}\)

Vì \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)

     \(\frac{1}{3^2}< \frac{1}{2\cdot3}\)

....

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

=> A < \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{\left(n-1\right)n}\)

=> A < \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)

=> A < \(1-\frac{1}{n}< 1\)(ĐPCM )

Vậy A < 1

Chững minh sao bạn !!!!!!!!!!!

gọi phân số đó là a/5 (a>0) 

vì 2/7<a/5<3/7 => 10/35<7a/35<15/35

=> 7a thuộc tập hợp 11, 12, 13, 14

ta có 35=7*5 => 7a= 14= 7*2 => 7a/35= 7*2/7*5= 2/5 ( thỏa mãn mẫu bằng 5)

kết luận..............

đặt b+c-a=x,a+c-b=y,a+b-c=z thì x,y,z>0 do a,b,c>0

=>x+y+z=a+b+c

có a=(y+z)/2 , b=(z+x)/2 ,c=(x+y)/2

A=(y+z)/2x + (z+x)/2y + (x+y)/2z =1/2[(x/y+y/x)+(y/z+z/y)+(x/z+z/x)

Áp dụng bđt cosi : x/y+y/x >= 2,y/z+z/y >= 2,z/x+x/z >= 2 

=>A >= 1/2.6=3 (đpcm)

Dấu "=" xảy ra <=> x=y=z<=>b+c-a=a+c-b=a+b-c<=>a=b=c <=> tam giác đó là tam gíac đều

Áp dụng bđt Cauchy-Schawrz dạng Engel ta có:

A = a^2/ab+ac-a^2  +  b^2/ab+bc-b^2  +  c^2/ac+bc-c^2

A \(\ge\)(a+b+c)^2/2.(ab+bc+ca)-(a^2+b^2+c^2)

A \(\ge\)a^2+b^2+c^2+2.(ab+bc+ca)/2.(ab+bc+ca)-(a^2+b^2+c^2)

A \(\ge\)2.(ab+bc+ca)-(a^2+b^2+c^2)/2.(ab+bc+ca)-(a^2+b^2+c^2)  +  2.(a^2+b^2+c^2)/2.(ab+bc+ca)-(a^2+b^2+c^2)

A \(\ge\)1  +  2.(a^2+b^2+c^2)/2.(a^2+b^2+c^2)-(a^2+b^2+c^2)

A \(\ge\) 1 + 2 = 3 (đpcm)

Dấu "=" xảy ra khi a = b = c

a: \(=\left(-\dfrac{25}{140}+\dfrac{245}{140}+\dfrac{32}{140}\right)\cdot\dfrac{-69}{20}\)

\(=\dfrac{252}{140}\cdot\dfrac{-69}{20}\)

\(=\dfrac{9}{5}\cdot\dfrac{-69}{20}=\dfrac{-621}{100}\)

b: \(=\left(6-2-\dfrac{4}{5}\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\)

\(=\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}=\dfrac{18}{5}\)

c: \(=\left(\dfrac{2}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)

\(=\dfrac{34}{24}\cdot\dfrac{-8}{17}=\dfrac{-1}{3}\cdot2=-\dfrac{2}{3}\)

a/ \(\frac{2}{3}+\frac{4}{35}< \frac{x}{105}< \frac{1}{7}+\frac{2}{5}+\frac{1}{3}\)

\(\Rightarrow\frac{82}{105}< \frac{x}{105}< \frac{92}{105}\)

\(\Rightarrow82< x< 92\)

\(\Rightarrow x=\left\{83;84;85;86;87;88;89;90;91\right\}\)

b/ \(-\frac{7}{15}+\frac{8}{60}+\frac{24}{90}\le\frac{x}{15}\le\frac{3}{5}+\frac{8}{30}+-\frac{4}{10}\)

\(\Rightarrow-\frac{1}{15}\le\frac{x}{15}\le\frac{7}{15}\)

\(\Rightarrow-1\le x\le7\)

\(\Rightarrow x=\left\{-1;0;1;2;3;4;5;6;7\right\}\)