Bài 2:
1) Ta có: A= 1+2+2^2+2^3+...+2^2020
2A= 2+2^2+2^3+2^4+...+2^2021
2A-A= (2+2^2+2^3+2^4+...+2^2021)- (1+2+2^2+2^3+...+2^2020)
A= 2^2021- 1
Mà 2^2021- 1< 2^2021
=> A<B
2) Ta có: 54^4= (18^3)^4= 18^12
Mà 18^12< 21^12
=> A< B
Bài 1:
7)(4x^2+1)^2020+ |y^2- 256|= 0(1)
Vì (4x^2+1)^2020 > 0 ;|y^2- 256| > 0
Từ (1)=>$\left \{ {{4x^2+1=0} \atop {|y^2- 256|=0}} \right.$ <=>$\left \{ {{4x^2=0+1} \atop {y^2- 256=0}} \right.$ <=> $\left \{ {{4x^2=1} \atop {y^2=0+256}} \right.$ <=>$\left \{ {{4x^2=1^2} \atop {y^2=256}} \right.$ <=>$\left \{ {{4x=1} \atop {y^2=16^2}} \right.$ <=>$\left \{ {{x=1:4} \atop {y=16}} \right.$ <=>$\left \{ {{x=$\frac{1}{4}$ } \atop {y=16}} \right.$
