A = 31.35 = 31.(33 + 2) = 31.33 + 62

B = 33.33 = 33.(31 + 2) = 31.33 + 66

B - A = 31.33 + 66 - 31.33 + 62 = 4 \(\Rightarrow\) B > A

A= 31 x35 = 31x (33+2)= 31x33+31x2

B = 33 x 33 = 33 x (31 + 2) = 33 x 31 + 33 x 2

Tao thay 31 x 33 + 31 x2 <  33x31 + 33 x 2

Suy ra A<B

tick cho anh cai nao

A = 33*33 = (31+2)*33 = 33*31 + 2*33 (1)

B = 31*35 = 31*(33+2) = 33*31 + 2*31 (2)

2*33 > 2*31 --> (1) > (2) hay A > B

Ta có : a-1/a = a/a - 1/a = 1 - 1/a  < 1

             b+1/b = b/b + 1/b = 1 + 1/b >1

=>  a-1/a < 1 < b+1/b

=> a-1/a < b+1/b

k mình nha 

dung rui

b.\(B=\dfrac{2n+5}{n+3}\)

\(B=\dfrac{n+n+3+3-1}{n+3}=\dfrac{n+3}{n+3}+\dfrac{n+3}{n+3}-\dfrac{1}{n+3}\)

\(B=1+1-\dfrac{1}{n+3}\)

Để B nguyên thì \(\dfrac{1}{n+3}\in Z\) hay \(n+3\in U\left(1\right)=\left\{\pm1\right\}\)

*n+3=1 => n=-2

*n+3=-1  => n= -4

Vậy \(n=\left\{-2;-4\right\}\) thì B có giá trị nguyên

3.

A:

2003²⁰⁰³+1=2003²⁰⁰².2003+1=2003²⁰⁰²+1

2003²⁰⁰⁴+1=2003²⁰⁰².2003.2003+1=2003²⁰⁰².2003+1(loại số 2003 thứ hai của cả mẫu số và tử số)  

B:

2003²⁰⁰²+1=2003²⁰⁰²+1

2003²⁰⁰³+1=2003²⁰⁰².2003+1

Suy ra: A=B

bằng nhau                               

Ta có: \(\frac{a}{b}=\frac{a.\left(b+1\right)}{b.\left(b+1\right)}=\frac{ab+a}{b.\left(b+1\right)}\)

          \(\frac{a+1}{b+1}=\frac{b.\left(a+1\right)}{b.\left(b+1\right)}=\frac{ab+b}{b.\left(b+1\right)}\)

Xét a>b

=>\(\frac{ab+a}{b.\left(b+1\right)}>\frac{ab+b}{b.\left(b+1\right)}\)

=>\(\frac{a}{b}>\frac{a+1}{b+1}\)

Xét a<b

=>\(\frac{ab+a}{b.\left(b+1\right)}