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

A=1.2+2.3+3.4+.............+2019.2020

3A=1.2.3+2.3.3+3.4.3+........................+2019.2020.3

3A=1.2.3+2.3.(4-1)+3.4.(5-2)+..............+2019.2020.(2021-2018)

3A=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+.............-2018.2019.2020+2019.2020.2021

3A=2019.2020.2021

A=\(\frac{2019.2020.2021}{3}\)

A=2747468660

Vậy A=2747468660

Chúc bn học tốt

\(A=1.2+2.3+3.4+.......+2019.2020\)

\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+......+2019.2020.3\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+.........+2019.2020.\left(2021-2018\right)\)

\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.......+2019.2020.2021-2018.2019.2020\)

\(=2019.2020.2021\)

\(\Rightarrow A=\frac{3A}{3}=\frac{2019.2020.2021}{3}=2747468660\)

Vậy \(A=2747468660\)

A=1.2+2.3+3.4+.............+2019.2020 3A=1.2.3+2.3.3+3.4.3+........................+2019.2020.3 3A=1.2.3+2.3.(4-1)+3.4.(5-2)+..............+2019.2020.(2021-2018) 3A=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+.............-2018.2019.2020+2019.2020.2021 3A=2019.2020.2021 A= 3 2019.2020.2021 A=2747468660 Vậy A=2747468660 Chúc bn học tốt

Sửa đề \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)

Ta có: \(\frac{2019}{1}+\frac{2018}{2}+...+\frac{1}{2019}\)

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

\(=2020+\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}-2020\)

\(=\frac{2020}{2}+...+\frac{2020}{2019}+\frac{2020}{2020}\)

\(=2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)\)Thay vào biểu thức A ta được:

\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}}{2020.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2020}\right)}=\frac{1}{2020}\)

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

giá trị biểu thức là 174

Ta có x = 2018

=> x + 1 = 2019

\(x^5-2019.x^4+2019.x^3-2019.x^2+2019.x-2020\)

\(=x^5-\left(x+1\right).x^4+\left(x+1\right).x^3-\left(x+1\right).x^2+\left(x+1\right).x-2020\)

\(=x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-2020\)

\(=x-2020\)

Thay x = 2018 vào biểu thức , ta được

\(2018-2020=-2\)

Vậy giá trị biểu thức là -2

Đặt \(x^2+y^2=a\)

Khi đó ta được: \(P=\left(a+2\right)^3-\left(a-2\right)^3-12a^2\)

\(\Leftrightarrow P=a^3.6a^2+12a+8-a^3+6a^2-12a+8-12a^2\)

\(\Leftrightarrow P=\left(a^3-a^3\right)+\left(6a^2+6a^2-12a^2\right)+\left(12a-12a\right)+8+8\)

\(\Leftrightarrow P=16\)

Vậy \(P=16\) tại \(x=2019\) và \(y=2020\)

+) \(M=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{2019\cdot2020}\)

\(M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2019}-\frac{1}{2010}\)

\(M=1-\frac{1}{2010}=\frac{2009}{2010}\)

Vậy M=\(\frac{2009}{2010}\)

+) Đặt A=\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\frac{1}{50}\right)\)

\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\cdot\cdot\cdot\frac{49}{50}\)

\(A=\frac{1\cdot2\cdot\cdot\cdot\cdot49}{2\cdot3\cdot\cdot\cdot\cdot50}=\frac{1}{50}\)

\(C=1-2+2^2-2^3+...-2^{2011}+2^{2012}\)

\(\Rightarrow2C=2-2^2+2^3-2^4+...-2^{2012}+2^{2013}\)

\(\Rightarrow3C=1+2^{2013}\)

\(\Rightarrow C=\frac{1+2^{2013}}{3}\)

Vậy 

\(D=-2+2^2-2^3+2^4-...-2^{2019}+2^{2020}\)

\(\Rightarrow-2D=2^2-2^3+2^4-2^5+...+2^{2020}-2^{2021}\)

\(\Rightarrow-3D=-2^{2021}+2\)

\(\Leftrightarrow D=\frac{2^{2021}-2}{3}\)

D \(=-2+2^2-2^3+2^4-...-2^{2019}+2^{2020}\)

\(\Leftrightarrow2D=-2^2+2^3-2^4+2^5-...-2^{2020}+2^{2021}\)

\(\Leftrightarrow2D+D=2^{2021}-2\)

\(\Leftrightarrow3D=2^{2021}-2\)

\(\Leftrightarrow D=\frac{2^{2021}-2}{3}\)

Vậy \(D=\frac{2^{2021}-2}{3}\)

@@ Học tốt

gọi a/2019=b/2020=c/2021 là x

\(\Rightarrow\)a=2019*x ;b=2020*x;c=2021*x

\(\Rightarrow\)M=4*(2019*x-2020*x)*(2020-2021)-(2021*x-2019*x)^2

\(\Rightarrow\)M=4*(-x)*(-x)-(2x)^2

\(\Rightarrow\)M=4*x^2-4*x^2

⇒M=0