Giải phương trình sau

(2x-3)² (x-4) =0

(x-2)(7-3x)=(x-2)(4x-3)

Đáp án`+`Giải thích các bước giải:

`(2x - 3)^2 (x - 4) = 0`

$\left[\begin{matrix} (2x - 3)^2 = 0\\ x - 4 = 0\end{matrix}\right.$

$\left[\begin{matrix} 2x - 3 = 0\\ x = 4\end{matrix}\right.$

$\left[\begin{matrix} x = \dfrac{3}{2}\\ x = 4\end{matrix}\right.$

Vậy `x ∈ {3/2;4}`

`(x - 2) (7 - 3x) = (x - 2) (4x - 3)`

`(x - 2) (7 - 3x) - (x - 2) (4x - 3) = 0`

`(x - 2) (7 - 3x - 4x + 3) = 0`

`(x - 2) (10 - 7x) = 0`

$\left[\begin{matrix} x - 2 = 0\\ 10 - 7x = 0\end{matrix}\right.$

$\left[\begin{matrix} x = 2\\ x = \dfrac{10}{7}\end{matrix}\right.$

Vậy `x ∈ {2;10/7}`

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{\text{a}}$$\color{#8c4ebd}{\text{t}}$$\color{#B592D6}
{\text{h}}$$\color{#ceaedf}{\text{s}}$