\(A=\frac{1990^{10}+1}{1990^{11}+1};B=\frac{1990^{11}+1}{1990^{12}+1}\)

Ta có: 

\(A=\frac{10\cdot\left(1990^{10}+1\right)}{10\cdot\left(1990^{11}+1\right)}\)

\(\Rightarrow A=\frac{1990^{11}+10}{1990^{12}+10}\)

\(\Rightarrow A=\frac{1990^{11}+1+9}{1990^{12}+1+9}\)

\(\Rightarrow A< B\)

Sửa đề: \(1990^{10}+1990^9\)

Ta có: \(1990^{10}+1990^9\)

\(=1990^9\cdot\left(1990+1\right)\)

\(=1990^9\cdot1991<1991^9\cdot1991=1991^{10}\)

Lời giải:

$A=1990^{10}+1990^9=1990^9(1990+1)=1990^9.1991< 1991^9.1991=1991^{10}$

Hay $A< B$

1990¹⁰ + 1990⁹  = 1990⁹ ( 1990 + 1 ) = 1990⁹ .1991

1991¹⁰ = 1991⁹ . 1991

-> 1990⁹ . 1991 < 1991⁹ . 1991 

Vậy  1990¹⁰ + 1990⁹ < 1991¹⁰

Tk cho mk nếu đúng nhé

Ta co : 1990¹⁰ + 1990⁹(1990+1)=1990⁹*1991

1991¹⁰=199⁹*1991

=> 1990⁹*1991<1991⁹ * 1991

Vay 1990¹⁰+ 1990⁹<1991¹⁰

ta có A= 1990^10+1990^9

suy ra A=1990^9 . ( 1990 + 1) = 1990^9  . 1991 mà ta có B= 1991^10 = 1991^9 . 1991

vì 1990^9 < 1991^9 suy ra A<B.chú ý dấu" . "  là dấu nhân

có cách nào rõ hơn ko

a, 202²⁰³=(101.2)²⁰³

=101²⁰³.2²⁰³

=101²⁰².2²⁰².202

b, 203²⁰²=(101,5.2)²⁰²

=101,5²⁰².2²⁰²

còn lại dễ

b, 1990¹⁰+1990⁹=1990⁹.1990+1990⁹=1990⁹.(1990+1)=1990⁹.1991

1991²⁰=1991¹⁹.1991

=>1990¹⁰+1990⁹<1991²⁰

c, 11¹⁹⁷⁹<11¹⁹⁸⁰=(11³)⁶⁶⁰=1331⁶⁶⁰

37¹³²⁰=(37²)⁶⁶⁰=1369⁶⁶⁰

=>11¹⁹⁷⁹<37¹³²⁰

quá dễ