\(A=2018\times2020+2021\) và \(B=2019\times2019+2021\)

\(A=2018\times2019+2018+2021\)

\(B=2018\times2019+2019+2021\)

Vì \(2019>2018\Rightarrow A< B\)

Ta có :

2018 x 2020 = 2018 x ( 2019 + 1 ) = 2018 + 2018 x 2019 < 2019 + 2018 x 2019 = 2019 x ( 2018 + 1 )

= 2019 x 2019

=> 2018 x 2020 < 2019 x 2019

=> 2018 x 2020 + 2021 < 2019 x 2019 + 2021

=> A < B

Bằng nhau

Ht cho mik 

Trả lời:

\(A=\frac{2}{2018.2020}+\frac{2021}{2020}-\frac{2020}{2019}\)

\(A=\frac{1}{2018}-\frac{1}{2020}+1+\frac{1}{2020}-\left(1+\frac{1}{2018}\right)\)

\(A=\frac{1}{2018}-\frac{1}{2020}+1+\frac{1}{2020}-1-\frac{1}{2018}\)

\(A=0\)

\(A=\frac{2}{2018}\cdot2020+\frac{2021}{2020}-\frac{2019}{2018}\)

\(A=\frac{2\cdot2020-2019}{2018}+\frac{2021}{2020}\)

\(A=\frac{2021}{2018}+\frac{2021}{2020}\)

\(A=\frac{2021\cdot\left(2020+2018\right)}{2018\cdot2020}=\frac{2021\cdot4038}{2018\cdot2020}=\frac{2021\cdot2019\cdot2}{2018\cdot1010\cdot2}=\frac{2020^2-1}{2018\cdot101\cdot10}\)

\(A=\frac{4080399}{20200180}\)

Trả lời

So sánh cái nào vs cái nào ạ

sao chỉ thấy có 1 vế ạ !

vi 2018/2019<1

   2019/2020<1

   2020/2021<1

nen 2018/2019 + 2019/2020 + 2020/2021<1+1+1=3

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

\(\left(2020\frac{2018}{2021}-2019\frac{20182018}{20212021}\right):\frac{2018}{2021}\)

\(=\left(2020\frac{2018}{2021}-2019\frac{2018}{2021}\right):\frac{2018}{2021}\)

\(=1:\frac{2018}{2021}=\frac{2021}{2018}\)

\(\left(2020\frac{2018}{2021}-2019\frac{20182018}{20212021}\right)\div\frac{2018}{2021}\)

\(=\left(2020\frac{2018}{2021}-2019\frac{2018}{2021}\right)\div\frac{2018}{2021}\)

\(=1\div\frac{2018}{2021}\)

\(=\frac{2021}{2018}\)

\(\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2019}x2018\)
\(=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{2018}{2019}=2-\dfrac{2019}{2018}=\dfrac{2017}{2018}\)