( 2x - 3 ) ( 6 - 2x ) = 0

+ TH1 : 2x -3 = 0 + TH2 : 6 - 2x = 0

2x - 3 = 0 6 - 2x = 0

2x = 0+ 3 2x = 6 - 0

2x = 3 2x = 6

x = 3 : 2 x = 6 : 2

x = \(\dfrac{3}{2}\) x = 3

vậy x = \(\dfrac{3}{2}\) hoặc x = 3

\(x^3-2x^2+2x=0\)

\(\Rightarrow x\left(x^2-2x+2\right)=0\)

\(\Rightarrow x\left(x^2-2.x.1+1^2+1\right)=0\)

\(\Rightarrow x\left[\left(x-1\right)^2+1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=-1\end{matrix}\right.\Rightarrow x=0\)

\(2x^3-5x^2+8x-5=0\)

\(\Rightarrow2x^3-3x^2-2x^2+5x+3x-5=0\)

\(\Rightarrow\left(2x^3-2x^2\right)-\left(3x^2-3x\right)+\left(5x-5\right)=0\)

\(\Rightarrow2x^2\left(x-1\right)-3x\left(x-1\right)+5\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(2x^2-3x+5\right)\)

Đề bài hơi căng,lẽ ra cái phương trình phải là x² + 2x chứ

em mới lớp 6 thui :( 

ảnh ko theo trật tự và bị thiếu nên mk sẽ gửi lại 1 tấm nx và mong bn thông cảm cho ​

Lập bảng xét dấu đii

2x(x-3) + 2x - 6 = 0

<=> 2x(x-3) + 2(x - 3) = 0

<=> (x-3)(2x+2)=0

<=> \(\left[{}\begin{matrix}x-3=0\\2x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

vậy............

Tìm x biết

2x.(x - 3) + 2x - 6 = 0

2x(x - 3) + 2(x - 3) = 0

(x - 3) (2x + 2) = 0

Suy ra: x - 3 = 0 hoặc 2x + 2 = 0

➢x - 3 = 0 _________ ➢2x + 2 = 0

⇒ x = 3 _________ ⇒ x = -1

Vậy x ∈ \(\left\{3;-1\right\}\)

a/ \(3x\left(x-2\right)-x+2=0\)

\(\Rightarrow3x\left(x-2\right)-\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(3x-1\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
b/ \(4x\left(x-3\right)-2x+6=0\)

\(\Rightarrow4x\left(x-3\right)-\left(2x-6\right)=0\)

\(\Rightarrow4x\left(x-3\right)-2\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(4x-2\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-3=0\\4x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)

c/ \(2x\left(x-4\right)+x-4=0\)

\(\Rightarrow\left(x-4\right)\left(2x+1\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-4=0\\2x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\x=-\dfrac{1}{2}\end{matrix}\right.\)

d/ \(2x^3+4x=0\)

\(\Rightarrow x\left(2x^2+4\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=0\\2x^2+4=0\Rightarrow x^2=-\dfrac{4}{2}=-2\end{matrix}\right.\)

Vì \(x^2=-2\) nên không xác định được x

Vậy x = 0

e/ \(3x^3-6x=0\)

\(\Rightarrow x\left(3x^2-6\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=0\\3x^2-6=0\Rightarrow x^2=\dfrac{6}{3}=2\Rightarrow\left\{{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)

a) Ta có: 4(x-5)+6(x-7)=0

⇔4x-20+6x-42=0

⇔10x-62=0

⇔10x=62

⇔\(x=\frac{31}{5}\)

Vậy: \(x=\frac{31}{5}\)

b) Ta có: 3(2x-7)-4(2x+3)=0

⇔6x-21-8x-12=0

⇔-2x-33=0

⇔-2x=33

⇔\(x=-\frac{33}{2}\)

Vậy: \(x=-\frac{33}{2}\)