2020x2019-1/2018x2020+2019=8238329618

\(\frac{2020.2019-1}{2018.2020+2019}\)

= \(\frac{2020.\left(2018+1\right)-1}{2018.2020+2019}\)

= \(\frac{2020.2018+2020-1}{2018.2020+2019}\)

= \(\frac{2020.2018+2019}{2018.2020+2019}\)

= \(1\)

Chúc bạn học tốt !!!!!

Rất vui vì giúp được bạn !!

\(\frac{2016}{2017}\)x \(\frac{2017}{2018}\)x \(\frac{2019}{2020}\)=\(\frac{504}{505}\)

đ/s:\(\frac{504}{505}\)

\(\frac{2016}{2017}\cdot\frac{2017}{2018}\cdot\frac{2018}{2019}\cdot\frac{2019}{2020}=\frac{504}{505}\)

504/505

\(\frac{2016}{2017}\times\frac{2017}{2018}\times\frac{2018}{2019}\times\frac{2019}{2020}\)=

\(0,998109801980198\)

Đổi ra ta sẽ có !

\(\frac{504}{505}\)

Vậy là  : ...................

= (1-1/2018)-(1+1/2018)-2020/2019

= 1-1/2018-1-1/2018-2020/2019

= -2/2018-2020/2019

vậy thôi

=(1-1/2018)-(1+1/2018)-2020/2019

=1-1/2018-1-1/2018-2020/2019

=-2/2018-2020/2019

`(-2/3)^2018*(3/2)^2020`

`=(2/3)^2018*(3/2)^2018*(3/2)^2`

`=(2/3*3/2)^2018*9/4`

`=1*9/4=9/4`

\(\left(\dfrac{-2}{3}\right)^{2018}\).\(\left(\dfrac{3}{2}\right)^{2020}\)=\(\left(\dfrac{-2}{3}\right)^{2018}\).\(\left(\dfrac{3}{2}\right)^{2018}\).\(\left(\dfrac{3}{2}\right)^2\)=\(\left(\dfrac{-2}{3}.\dfrac{3}{2}\right)^{2018}\).\(\dfrac{9}{4}\)=-1.\(\dfrac{9}{4}\)=\(\dfrac{-9}{4}\)

Tham khảo:

-\(\dfrac{2}{3}\)²⁰¹⁸\(\times\)\(\dfrac{3}{2}\)²⁰²⁰

=-\(\dfrac{2^{2018}\times3^{2020}}{3^{2018}\times2^{2020}}\)

=-\(\dfrac{3^2}{2^2}\)=-\(\dfrac{9}{4}\)

Chúc bạn học tốt.

Ta có: \(\left(x-2\right)^{2020}\ge0\forall x\)

\(\left(y+5\right)^{2018}\ge0\forall y\)

Do đó: \(\left(x-2\right)^{2020}+\left(y+5\right)^{2018}\ge0\forall x,y\)

Dấu '=' xảy ra khi (x,y)=(2;-5)

a,1

b,1

A và B đáp số 1

a,b đều là 1

x = 2020 => 2021 = x + 1

x²⁰²⁰ - 2021x²⁰¹⁹ + 2021x²⁰¹⁸ - 2021x²⁰¹⁷ + ... + 2021x² - 2021x + 1

= x²⁰²⁰ - ( x + 1 )x²⁰¹⁹ + ( x + 1 )x²⁰¹⁸ - ( x + 1 )x²⁰¹⁷ + ... + ( x + 1 )x² - ( x + 1 )x + 1

= x²⁰²⁰ - x²⁰²⁰ - x²⁰¹⁹ + x²⁰¹⁹ + x²⁰¹⁸ - x²⁰¹⁸ - x²⁰¹⁷ + ... + x³ + x² - x² - x + 1

= -x + 1 = -2020 + 1 = -2019

Vậy giá trị của biểu thức = -2019