A=11+22+33+44

=33+77

=110

B=14+23+32+41

=37+73

=110

VÌ 110=110 NÊN A =B

A = 11 + 22 + 33 + 44

   = 33 + 33 + 44

   = 66 + 44

   = 110

B = 14 + 23 + 32 + 41

   = 37 + 73

   = 110

Vậy A = B ( = 110 )

\(A=11+22+33+44=10+1+20+2+30+3+40+4\)

\(A=10+4+20+3+30+2+40+1=14+23+32+41=B\)

ta có x^100> 0

=> x>0 

mà x^100=x

suy ra x^99=1  ( chia cả 2 vế cho x )

=> x=1

3^22=3^21.3
2^33=2^31.4=2^31.(3+1)=2^31.3+2^31
Ta có 3^22=(3^2)^11=9^11
2^33=(2^3)^11=8^11
→2^31.3+2^31<3^21.3→2^31.3<3^21.3→2^31<3^21

Ta co : 22³³ va 33²²

=> 22³³=22^3.11=(22³)¹¹=10648¹¹

=> 33²²=33^2.11=(33²)¹¹=1089¹¹

Ma 1089¹¹<10648¹¹

Vay suy ra 22³³<33²²

^{nho lik e} 

Mất công chép lại:

3^22=3^21.3
2^33=2^31.4=2^31.(3+1)=2^31.3+2^31
Ta có 3^22=(3^2)^11=9^11
2^33=(2^3)^11=8^11
→2^31.3+2^31<3^21.3→2^31.3<3^21.3→2^31<3^21

\(\dfrac{22}{33}=\dfrac{22:11}{33:11}=\dfrac{2}{3}\)
Ta có: 
Mẫu số chung 2 phân số: 57
\(\dfrac{12}{19}=\dfrac{12*3}{19*3}=\dfrac{36}{57}\)
\(\dfrac{2}{3}=\dfrac{2*19}{3*19}=\dfrac{38}{57}\)
Vì \(36< 38\) nên\(\dfrac{36}{57}< \dfrac{38}{57}\)
Vậy \(\dfrac{12}{19}< \dfrac{22}{33}\)

\(\dfrac{22}{33}\)=\(\dfrac{22:11}{33:11}\)=\(\dfrac{2}{3}\)=\(\dfrac{2\times6}{3\times6}\)=\(\dfrac{12}{18}\)>\(\dfrac{12}{19}\)

Ta có : \(\dfrac{1}{2}< \dfrac{1}{1.2};\dfrac{1}{2^2}< \dfrac{1}{2.3};...;\dfrac{1}{2^{10}}< \dfrac{1}{9.10}\)

\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}=\dfrac{9}{10}< 1\Rightarrow A< B\)

Ta có \(3^{22}=\left(3^2\right)^{11}=9^{11}\)

\(2^{33}=\left(2^3\right)^{11}=8^{11}\)

Ta có \(8< 9\Rightarrow8^{11}< 9^{11}\)

hay \(3^{22}>2^{33}\)

Ta có 3^22= (3^2)^11=9^11(1)

2^33= (2^3)^11=8^11(2)

Từ (1) và (2) suy ra 3^22>2^33

Học tốt !

mình nghĩ là 2^33 < 3^22

nấu sai thì bạn thông cảm nhé :)

Ta có:2^33=(2^3)^11=6^11

      3^22=(3^2)^11=6^11

:) 2^33=3^22

ta có: 2\(^{33}\)= (2\(^3\))\(^{11}\)=8\(^{11}\)

3\(^{22}\)=(3\(^2\))\(^{11}\)= 9\(^{11}\)

Do 9>8 ⇒ 9\(^{11}\)> 8\(^{11}\) hay 3\(^{22}\)>2\(^{33}\)

a: \(33^{44}=\left(33^4\right)^{11}\)

\(44^{33}=\left(44^3\right)^{11}\)

mà \(33^4>44^3\)

nên \(33^{44}>44^{33}\)