Tìm x biết 4.(x+1)+3.(x+3)=7 x^2-3x=0 2020x.(x-2019)+x_2019=0

Đáp án+Giải thích các bước giải:

`4(x+1)+3(x+3)=7`

`<=> 4x+4+3x+9=7`

`<=> 7x+13=7`

`<=> 7x=-6`

`<=> x=-6/7`

Vậy `x=-6/7`

`x^2-3x=0`

`<=> x(x-3)=0`

`<=> [(x=0),(x-3=0):}`

`<=> [(x=0),(x=3):}`

Vậy `x \in {0;3}`

`2020x(x-2019)+x-2019=0`

`<=> 2020x(x-2019)+(x-2019)=0`

`<=> (2020x+1)(x-2019)=0`

`<=> [(2020x+1=0),(x-2019=0):}`

`<=> [(x=-1/2020),(x=2019):}`

Vậy `x \in {-1/2020;2019}` 

Answer

`4 . (x + 1) + 3 . (x + 3) = 7`

`=> 4 . x + 4 . 1 + 3 . x + 3 . 3 = 7`

`=> 4x + 4 + 3x + 9 = 7`

`=> (4x + 3x) + (4 + 9) = 7`

`=> 7x + 13 = 7`

`=> 7x = 7 - 13`

`=> 7x = -6`

`=> x = -6 : 7`

`=> x = -6/7`

Vậy `S = {-6/7}

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`x^2 - 3x = 0`

`=> x . (x - 3) = 0`

`=>` $\left[\begin{matrix} x = 0\\ x - 3 = 0\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 0\\ x = 0 + 3\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = 0\\ x = 3\end{matrix}\right.$

Vậy `S = {0 ; 3}`

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`2020x . (x - 2019) + x - 2019 = 0`

`=> 2020x . (x - 2019) + (x - 2019) = 0`

`=> (2020x + 1) . (x - 2019) = 0`

`=>` $\left[\begin{matrix} 2020x + 1 = 0\\ x - 2019 = 0\end{matrix}\right.$

`=>` $\left[\begin{matrix} 2020x = 0 - 1\\ x = 0 + 2019\end{matrix}\right.$

`=>` $\left[\begin{matrix} 2020x = - 1\\ x = 2019\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = - 1 : 2020\\ x = 2019\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = \dfrac{-1}{2020}\\ x = 2019\end{matrix}\right.$

Vậy `S = {-1/2020 ; 2019}`