oh no bài thứ nhất là dạng chứng minh cs đúng ko ,

ko thể nào là dạng tìm a,b,c đc-.-

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{2022}\)

hay \(\dfrac{ab+bc+ca}{abc}=\dfrac{1}{a+b+c}\)

\(\Leftrightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)

\(\Leftrightarrow a^2b+ab^2+b^2c+bc^2+c^2a+ca^2+3abc=abc\)

\(\Leftrightarrow a^2b+ab^2+b^2c+bc^2+c^2a+ca^2+2abc=0\)

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

-Xét a + b = 0 => P = 2022^2021

Bạn xét tương tự với b + c = 0 và c + a = 0 dc P = 2022^2021 nhé

a+bab+a+bc(a+b+c)=0a+bab+a+bc(a+b+c)=0

(a+b)[ab+bc+ca+c2abc(a+b+c)]=0(a+b)[ab+bc+ca+c2abc(a+b+c)]=0

 ⇔ a=−b

 ⇔ b=−c

 ⇔ c=−a

Thay vào P từng cái rồi tính tiếp nhé

a) Ta có:

Suy ra: 

Do đó .

Lại có .

Vậy A < B.

a: \(98^{10}\cdot A=\dfrac{98^{98}+98^{10}}{98^{98}+1}=1+\dfrac{98^{10}-1}{98^{98}+1}\)

\(98^{10}\cdot B=\dfrac{98^{99}+98^{10}}{98^{99}+1}=1+\dfrac{98^{10}-1}{98^{99}+1}\)

98^88+1>98^99+1

=>A<B

b: \(\dfrac{1}{2022^2}\cdot C=\dfrac{2022^{2023}+1}{2022^{2023}+2022^2}=1+\dfrac{1-2022^2}{2022^{2023}+2022^2}\)

\(\dfrac{1}{2022^2}\cdot D=\dfrac{2022^{2021}+1}{2022^{2021}+2022^2}=1+\dfrac{1-2022^2}{2022^{2021}+2022^2}\)

2022^2023>2022^2021

=>2022^2023+2022^2>2022^2021+2022^2

=>\(\dfrac{2022^2-1}{2022^{2023}+2022^2}< \dfrac{2022^2-1}{2022^{2021}+2022^2}\)

=>\(\dfrac{1-2022^2}{2022^{2023}+2022^2}>\dfrac{1-2022^2}{2022^{2021}+2022^2}\)

=>C>D

\(A=\dfrac{3\cdot\dfrac{a}{b}-\dfrac{-a}{b}}{-\dfrac{-5a}{b}+\dfrac{4a}{b}}\\ =\left(\dfrac{3a}{b}+\dfrac{a}{b}\right):\left(\dfrac{5a}{b}+\dfrac{4a}{b}\right)\\ =\dfrac{4a}{b}:\dfrac{9a}{b}\\ =\dfrac{4a}{b}\cdot\dfrac{b}{9a}\\ =\dfrac{4}{9}\)

Vậy `a=2021/2022` ; `b=2023/2022` thì `A=4/9`

\(B=\left(\dfrac{2020}{2}+1\right)+\left(\dfrac{2019}{3}+1\right)+...+\left(\dfrac{1}{2021}+1\right)+1\)

\(=\dfrac{2022}{2}+\dfrac{2022}{3}+...+\dfrac{2022}{2021}+\dfrac{2022}{2022}\)

=2022(1/2+1/3+...+1/2021+1/2022)

=>B/A=2022

\(B=\dfrac{\dfrac{2ab}{3}-\dfrac{3ab}{2}}{-\dfrac{5bb}{6}}\)

\(=\dfrac{\dfrac{4ab}{6}-\dfrac{9ab}{6}}{-\dfrac{5bb}{6}}\)

\(=\dfrac{-\dfrac{5ab}{6}}{-\dfrac{5bb}{6}}=\dfrac{ab.\dfrac{5}{6}}{bb.\dfrac{5}{6}}\)

\(=\dfrac{ab}{bb}=\dfrac{a}{b}\)

Với \(a=\dfrac{2021}{2022};b=\dfrac{2023}{2022}\), ta được:

\(B=\dfrac{2021}{2022}:\dfrac{2023}{2022}=\dfrac{2021}{2022}.\dfrac{2022}{2023}=\dfrac{2021}{2023}\)

d

d

giải thích cho những ng ko hỉu ;-;

\(=\dfrac{\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{2022}}{\left(\dfrac{2016}{6}+1\right)+\left(\dfrac{2015}{7}+1\right)+...+\left(\dfrac{1}{2021}+1\right)+1}\)

\(=\dfrac{\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{2022}}{\dfrac{2022}{6}+\dfrac{2022}{7}+...+\dfrac{2022}{2021}+\dfrac{2022}{2022}}\)

\(=\dfrac{\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{2022}}{2022.\left(\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{2022}\right)}=\dfrac{1}{2022}\)

a) 2021 - (1/3)² . 3²

= 2021 - 1/9 . 9

= 2021 - 1

= 2020

b) 5/10 + 9 . (-3/2)

= 1/2 - 27/2

= -26/2

= -13

c) -10 . (-2021/2022)⁰ + (2/5)² : 2

= -10 . 1 + 4/25 . 2

= -10 + 8/25

= -68/7

\(a,2021-\left(\dfrac{1}{3}\right)^2\cdot3^2\\ =2021-\dfrac{1}{9}\cdot9\\ =2021-\dfrac{9}{9}\\ =2021-1=2020\\ b,\dfrac{5}{10}+9\cdot\dfrac{-3}{2}\\ =\dfrac{5}{10}+\dfrac{-27}{2}\\ =\dfrac{5}{10}+\dfrac{-135}{10}\\ =-\dfrac{130}{10}\\ =-13\\ c,-10\cdot\left(-\dfrac{2021}{2022}\right)^0+\left(\dfrac{2}{5}\right)^2:2\\ =-10\cdot1+\dfrac{4}{25}\cdot\dfrac{1}{2}\\ =-10+\dfrac{4}{50}\\ =-10+\dfrac{2}{25}\\ =-\dfrac{248}{25}\)