a) \(A=2019.2021=\left(2020-1\right).\left(2020+1\right)=2020^2-1\)

\(B=2020.2020=2020^2\)

\(\Rightarrow2020^2-1< 2020^2\)\(\Rightarrow A< B\)

b) \(C=35.53-18=\left(34+1\right).53-18=34.53+53-18=34.53+34\)

mà \(D=35+53.34\)

\(\Rightarrow C=D\)

18:

a: \(S=3\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{98\cdot100}\right)\)

=3*(1/2-1/4+1/4-1/6+...+1/98-1/100)

=3*49/100=147/100

b: Để A là số nguyên thì n-1 thuộc Ư(2)

=>n-1 thuộc {1;-1;2;-2}

=>n thuộc {2;0;3;-1}

a: \(\dfrac{-11}{18}+\dfrac{12}{29}+\dfrac{-7}{18}+\dfrac{2020}{2021}+\dfrac{17}{29}\)

=(-11/18-7/18)+(12/29+17/29)+2020/2021

=2020/2021

b: \(\dfrac{-2}{3}\cdot\dfrac{4}{9}+\dfrac{5}{9}\cdot\dfrac{-2}{3}+\dfrac{4}{3}\)

=-2/3+4/3=2/3

𝑝=−2856279824648840

a: \(17A=\dfrac{17^{19}+17}{17^{19}+1}=1+\dfrac{16}{17^{19}+1}\)

\(17B=\dfrac{17^{18}+17}{17^{18}+1}=1+\dfrac{16}{17^{18}+1}\)

mà 17^19+1>17^18+1

nên A<B

b: \(2C=\dfrac{2^{2021}-2}{2^{2021}-1}=1-\dfrac{1}{2^{2021}-1}\)

\(2D=\dfrac{2^{2022}-2}{2^{2022}-1}=1-\dfrac{1}{2^{2022}-1}\)

2^2021-1<2^2022-1

=>1/2^2021-1>1/2^2022-1

=>-1/2^2021-1<-1/2^2022-1

=>C<D

A = B

Tham khảo:

Đặt \(2020-x=u;x-2021=v\)thì \(u+v=-1\)

Phương trình trở thành \(\frac{u^2+uv+v^2}{u^2-uv+v^2}=\frac{19}{49}\Leftrightarrow30u^2+30v^2+68uv=0\)

\(\Leftrightarrow15\left(u+v\right)^2+4uv=0\Leftrightarrow4uv=-15\Leftrightarrow uv=\frac{-15}{4}\)

hay \(\left(2020-x\right)\left(x-2021\right)=-\frac{15}{4}\Leftrightarrow x^2-4041x+4082416,25=0\)

Dùng công thức nghiệm tìm được x = 2022, 5 hoặc x = 2018, 5

Giải:

Ta có: N=2019+2020/2020+2021

=>N=2019/2020+2021 + 2020/2020+2021

Vì 2019/2020 > 2019/2020+2021 ; 2020/2021 > 2020/2020+2021

=>M>N

Vậy ...

Chúc bạn học tốt!

Ta có : \(\dfrac{2019}{2020}>\dfrac{2019}{2020+2021}\)

            \(\dfrac{2020}{2021}>\dfrac{2020}{2020+2021}\)

\(\Rightarrow\dfrac{2019}{2020}+\dfrac{2020}{2021}>\dfrac{2019+2020}{2020+2021}\)

\(\Rightarrow M>N\)