a) 5x( x - 2021 ) - x + 2021 = 0

 5x( x - 2021 ) - ( x - 2021 ) = 0

 ( x - 2021 ) ( 5x - 1 ) = 0

⇔  x - 2021 = 0 ⇒ x = 2021

⇔ 5x - 1 =0 ⇒ x = $\frac{1}{5}$ 

Vậy x = 2021 hoặc x = $\frac{1}{5}$ 

b) ( x - 2 )² - ( x - 3 )( x + 3 ) = 6

 ( x - 2 )² - ( x² - 9 ) = 6

 x² - 4x + 4 - x² + 9 = 6

 13 - 4x = 6

4x = 7

x = $\frac{7}{4}$ 

Vậy x = $\frac{7}{4}$ 

c) x( x - 5 ) +  x - 5 = 0 

 x( x - 5 ) + ( x - 5 ) = 0

 ( x - 5 ) ( x +1 ) = 0

⇔ x - 5 = 0 ⇒ x =5

⇔ x + 1 = 0 ⇒  x = -1

Vậy x = 5 và x = -1

 

Đáp án:

`a,` `5x.(x - 2021) - x + 2021 = 0`

`↔  5x.(x - 2021) -  (x - 2021) = 0`

`↔ (x - 2021).(5x - 1) = 0`

`↔` $\left[\begin{matrix} x - 2021 = 0\\ 5x-1=0\end{matrix}\right.$

`↔` $\left[\begin{matrix} x = 2021\\ x = \dfrac{1}{5}\end{matrix}\right.$

Vậy `x ∈ {2021; 1/5}`

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`b,` `(x - 2)^2 - (x - 3).(x + 3) = 6`

`↔ x^2 - 4x + 4 - (x^2 - 3^2) = 6`

`↔ x^2 - 4x + 4 - x^2 + 9 = 6`

`↔ (x^2 - x^2) -4x + (4 + 9) = 6`

`↔ -4x + 13 = 6`

`↔ -4x = -7`

`↔ x = 7/4`

Vậy `x = 7/4`

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`c,` `x.(x - 5) + x - 5 = 0`

`↔ x.(x - 5) + (x - 5) = 0`

`↔ (x - 5).(x + 1) = 0`

`↔` $\left[\begin{matrix} x - 5 = 0\\ x + 1=0\end{matrix}\right.$

`↔` $\left[\begin{matrix} x = 5\\ x = -1\end{matrix}\right.$

Vậy `x ∈ {5; -1}`

`#dariana`