D

D

Ta có |x| \(\ge\) 0 \(\forall\) x

\(\Rightarrow\left|x\right|+2020\ge2020\)

D

bn tham khảo link này nhé

https://olm.vn/hoi-dap/tim-kiem?id=1300742&subject=1&q=++++++++++kh%C3%B4ng+t%C3%ADnh+k%E1%BA%BFt+qu%E1%BA%A3+c%E1%BB%A5+th%E1%BB%83+h%C3%A3y+so+s%C3%A1nh+a+v%C3%A0+b+++a=+2020+.+2020b=+2018+.+2022+++++++++

a>b

2020 x 2020 > 2019 x 2021

\(2019\times2021=\left(2020-1\right)\left(2020+1\right)=2020^2-1< 2020^2=2020\times2020\)

2020x2020>2019x2021

Ta có: \(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)

mà \(a+b+c\ne0\)

nên \(a^2+b^2+c^2-ab-ac-bc=0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Leftrightarrow a=b=c\)

Ta có: \(M=\dfrac{a^{2020}+b^{2020}+c^{2020}}{\left(a+b+c\right)^{2020}}\)

\(=\dfrac{a^{2020}+a^{2020}+a^{2020}}{\left(a+a+a\right)^{2020}}=\dfrac{3\cdot a^{2020}}{9\cdot a^{2020}}=\dfrac{1}{3}\)

Đáp án là a>b!

Học tốt!

a) A= 2018 x 2022 và B= 2020 x 2020

ta có :

A=2018.2022=2020.2022-2.2022

=2020.2020+2020.2-2.2022

=2020.2020+2(2020-2022)

=2020.2020-4=B-4

=>A=B-4

hay B > A 4 đơn vị

Ta có: $a=2020.2020=\left(2018+2\right).2020=2018.2020+2020.2$

          $b=2018.2020=2018.\left(2020+2\right)=2018.2020+2018.2$

Vì $2018.2020=2018.2020$

Mà $2020.2>2018.2$

Nên $2018.2020+2020.2>2018.2020+2018.2$

Vậy $a>b$

Ta có : a³ + b³ + c³ = 3abc

=> (a + b)(a² - ab + b²) + c³ - 3abc = 0

=> (a + b)³ - 3ab(a + b) + c³ - 3abc = 0

=> [(a + b)³ + c³] - [(3ab(a + b) + 3abc] = 0

=> (a + b + c)(a² + b² + 2ab - ac - bc + c²) - 3ab(a + b + c) = 0

=> (a + b + c)(a² + b² + c² - ab - ac - bc) = 0

=> a² + b² + c² - ab- ac - bc = 0

=> 2(a² + b² + c² - ab- ac - bc) = 0

=> 2a² + 2b² + 2c² - 2ab - 2ac - 2bc = 0

=> (a² - 2ab + b²) + (b² - 2bc + c²) + (a² - 2ac + c²) = 0

=> (a - b)² + (b - c)² + (a - c)² = 0

=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Rightarrow a=b=c\)

Khi đó M = \(\frac{a^{2020}+b^{2020}+c^{2020}}{\left(a+b+c\right)^{2020}}=\frac{3.c^{2020}}{\left(3c\right)^{2020}}+\frac{3c^{2020}}{3^{2020}.c^{2020}}=\frac{1}{3^{2019}}\)

a: Ta có: \(-\left(x+5\right)^2\le0\forall x\)

\(\Leftrightarrow-\left(x+5\right)^2+2021\le2021\forall x\)

Dấu '=' xảy ra khi x=-5

a ) 

Ta có : 

\(A=18\times19=\left(17+1\right)\times19=17\times19+19\)

\(B=17\times20=17\times\left(19+1\right)=17\times19+17\)

Do \(17\times19+19>17\times19+17\)

\(\Rightarrow A>B\)

Vậy \(A>B\)

b ) 

Ta có : 

\(C=2019\times2019=\left(2018+1\right)\times2019=2018\times2019+2019\)

\(D=2018\times2020=2018\times\left(2019+1\right)=2018\times2019+2018\)

Do \(2018\times2019+2019>2018\times2019+2018\)

\(\Rightarrow C>D\)

Vậy \(C>D\)

a) A < B

b) C > D

Học tốt

@@@

a ) Ta có :

A = 18 x 19 = 18 x ( 17 + 2 ) = 18 x 17 + 18 x 2

B = 17 x 20 = 17 x ( 18 + 2 ) = 17 x 18 + 17 x 2

MÀ 18 x 17 + 18 x 2 > 17 x 18 + 17 x 2 Do đó A > B

Vậy A > B

\(x=\dfrac{1}{\sqrt{2}}\left(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\right)=\sqrt{6}\)

\(y=\sqrt{\left(\sqrt{6}-1\right)^2}=\sqrt{6}-1\)

\(\Rightarrow x-y=1\Rightarrow P=1\)

\(B=x-2020-\sqrt{x-2020}+\dfrac{1}{4}+\dfrac{8079}{4}\)

\(B=\left(\sqrt{x-2020}-\dfrac{1}{2}\right)^2+\dfrac{8079}{4}\ge\dfrac{8079}{4}\)

\(B_{min}=\dfrac{8079}{4}\) khi \(x=\dfrac{8081}{4}\)