C

C

C

\(\dfrac{x-4}{2022}+\dfrac{x-3}{2021}+\dfrac{x-2}{2020}+\dfrac{x-1}{2019}\text{=}-4\)

\(\dfrac{x-4}{2022}+\dfrac{x-3}{2021}+\dfrac{x-2}{2020}+\dfrac{x-1}{2019}+4\text{=}0\)

\(\left(\dfrac{x-4}{2022}+1\right)+\left(\dfrac{x-3}{2021}+1\right)+\left(\dfrac{x-2}{2020}+1\right)+\left(\dfrac{x-1}{2019}+1\right)\text{=}0\)

\(\dfrac{x-2018}{2022}+\dfrac{x-2018}{2021}+\dfrac{x-2018}{2020}+\dfrac{x-2018}{2019}\text{=}0\)

\(\left(x-2018\right)\left(\dfrac{1}{2022}+\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}\right)\text{=}0\)

\(Do:\) \(\dfrac{1}{2022}+\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}\ne0\)

\(x-2018\text{=}0\)

\(x\text{=}2018\)

\(Vậy...\)

a, \(=\dfrac{1.2.3...5}{2.3...6}=\dfrac{1}{6}\)

\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}+\dfrac{x+3}{2018}+\dfrac{x+4}{2017}+4=0\)

⇔ \(\dfrac{x+1}{2020}+1+\dfrac{x+2}{2019}+1+\dfrac{x+3}{2018}+1+\dfrac{x+4}{2017}+1=0\)

\(\Leftrightarrow\) \(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}+\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}=0\)

⇔ \(\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}+\dfrac{1}{2017}\right)=0\)

\(Do\) \(\left(\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}+\dfrac{1}{2017}\right)\ne0\)

⇒ \(x+2021=0\)

⇔ \(x=-2021\)

\(Vậy\) \(x=-2021\)

a) (x - 1)³ - 1 = 0

<=> (x - 1)³ = 0 + 1

<=> (x - 1)³ = 1

<=> (x - 1)³ = 1³

<=> x - 1 = 1

<=> x = 1 + 1

<=> x = 2

=> x = 2

b) (x - 4)²⁰¹⁹ = 1

<=> (x - 4)²⁰¹⁹ = 1²⁰¹⁹

<=> x - 4 = 1

<=> x = 1 + 4

<=> x = 5

=> x = 5

c) (x - 2019)²⁰²⁰ = 0

<=> (x - 2019)²⁰²⁰ = 0²⁰²⁰

<=> x - 2019 = 0

<=> x = 0 + 2019

<=> x = 2019

=> x = 2019

d) (x - 1)² = (x - 1)³

<=> x² - 2x + 1 = x³ - 2x² + x - x² + 2x - 1

<=> x² - 2x + 1 = x³ - 3x² + 3 - 1

<=> x² - 2x + 1 - x³ + 3x² - 3 + 1 = 0

<=> 4x² - 5x + 2 - x³ = 0

<=> (-x² + 3x - 2)(x - 1) = 0

<=> (x² - 3x + 2)(x - 1) = 0

<=> (x - 2)(x - 1)(x - 1) = 0

<=> x - 2 = 0 hoặc x - 1 = 0

       x = 0 + 2         x = 0 + 1

       x = 2               x = 1

=> x = 1 hoặc x = 2

\(A=\left(2020\times2019+2019\times2018\right)\times\left(1+\dfrac{1}{2}:1\dfrac{1}{2}-1\dfrac{1}{3}\right)\)

\(A=\left[2019\times\left(2020+2018\right)\right]\times\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)\)

\(A=4038\times2019\times\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)\)

\(A=4038\times2019\times0\)

\(A=0\)

Ta có :\(\frac{x+4}{2018}+\frac{x+3}{2019}=\frac{x+2}{2020}+\frac{x+1}{2021}\)

=> \(\left(\frac{x+4}{2018}+1\right)+\left(\frac{x+3}{2019}+1\right)=\left(\frac{x+2}{2020}+1\right)+\left(\frac{x+1}{2021}+1\right)\)

=> \(\frac{x+2022}{2018}+\frac{x+2022}{2019}=\frac{x+2022}{2020}+\frac{x+2022}{2021}\)

=> \(\frac{x+2022}{2018}+\frac{x+2022}{2019}-\frac{x+2022}{2020}-\frac{x+2022}{2021}=0\)

=> \(\left(x+2022\right)\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\right)=0\)

Vì \(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\ne0\)

=> x + 2022 = 0

=> x = -2022

Vậy x = -2022

\(\frac{x+4}{2018}+\frac{x+3}{2019}=\frac{x+2}{2020}+\frac{x+1}{2021}\)  

\(\frac{x+4}{2018}+1+\frac{x+3}{2019}+1=\frac{x+2}{2020}+1+\frac{x+1}{2021}+1\) 

\(\frac{x+4}{2018}+\frac{2018}{2018}+\frac{x+3}{2019}+\frac{2019}{2019}=\frac{x+2}{2020}+\frac{2020}{2020}+\frac{x+1}{2021}+\frac{2021}{2021}\)   

\(\frac{x+2022}{2018}+\frac{x+2022}{2019}=\frac{x+2022}{2020}+\frac{x+2022}{2021}\)   

\(\frac{x+2022}{2018}+\frac{x+2022}{2019}-\frac{x+2022}{2020}-\frac{x+2022}{2021}=0\)   

\(\left(x+2022\right)\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\right)=0\)   

\(x+2022=0\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\ne0\right)\)   

\(x=0-2022\) 

\(x=-2022\)

A=3(x-4)⁴

Vì (x-4)⁴ ≥0

=>3(x-4)⁴ ≥0

Vậy MinA=0

B=5+2(x-2019)²⁰²⁰

Vì (x-2019)²⁰²⁰ ≥0

=>5+(x-2019)²⁰²⁰ ≥5

Để B đạt Min 

=>x-2019=0

=>x=2019

Vậy MinB=5 <=>x=2019