Đặt biểu thức đã cho là A

Đặt \(B=2^{2019}+2^{2018}+.......+2^1+2^0\)

\(\Rightarrow2B=2^{2020}+2^{2019}+.......+2^2+2\)

\(\Rightarrow2B-B=B=2^{2020}-2^0\)

\(\Rightarrow A=2^{2020}-\left(2^{2020}-2^0\right)=2^{2020}-2^{2020}+1=1\)

\(\frac{2^{2019}+2^{2020}}{2^{2019}+2^{2018}}\)

\(=\frac{2^{2019}.\left(1+2\right)}{2^{2018}.\left(2+1\right)}\)

\(=\frac{2^{2019}}{2^{2018}}=2\)

a: \(A=1-\dfrac{2\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}{4\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}\)

=1-2/4=1/2

b: \(B=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\dfrac{-6}{9}=-\dfrac{10}{3}\)

c: x-y=0 nên x=y

\(C=x^{2020}-x^{2020}+y\cdot y^{2019}-y^{2019}\cdot y+2019\)

=2019

Nhận xét : ( x + y - 3 )^2018 >=0 và 2018.(2x-4)^2020 >= 0

=> (x+y-3)^2018 + 2018.(2x-4)^2020 >=0 

Dấu = xảy ra khi : x + y - 3 = 0 và 2x - 4 = 0 => x = 2 và y = 1

Thay vào bt S :

S = ( 2 - 1)^2019 + (2-1)^2019

= 1^2019 + 1^2019 = 2

em cảm ơn

f(0)=2018

=>\(a\cdot0^2+b\cdot0+c=2018\)

=>c=2018

=>f(x)=\(ax^2+bx+2018\)

f(1)=2019

=>\(a\cdot1^2+b\cdot1+2018=2019\)

=>a+b=1

f(-1)=2020

=>\(a\cdot\left(-1\right)^2+b\cdot\left(-1\right)+2018=2020\)

=>a-b=2

mà a+b=1

nên a=(2+1)/2=1,5; b=a-2=1,5-2=-0,5

=>f(x)=1,5x^2-0,5x+2018

f(2)=1,5*2^2-0,5*2+2018

=6-1+2018

=2018+5

=2023

Ta có :

f(0) = a.0^2 + b.0 + c = 2018 => c = 2018

f(1) = a + b + c = 2019 => a + b = 1

f(-1) = a - b + c = 2020 => a - b = 2

Suy ra : a = 1,5 ; b = = - 0,5

Vậy : f(x) = 1,5x^2 - 0,5x + 2018

Suy ra: f(2) = 1,5.2^2 - 0,5.2 + 2018 = 2023