a) \(x\left(x-6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b) \(\left(-7-x\right)\left(-x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)

c) \(\left(x+3\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

d) \(\left(x-3\right)\left(x^2+12\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)

\(\Rightarrow x=3\)

e) \(\left(x+1\right)\left(2-x\right)\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)

\(\Rightarrow-1\le x\le2\)

f) \(\left(x-3\right)\left(x-5\right)\le0\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow3\le x\le5\)

a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)

d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3

c) => \(\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

Bài 2: 

a: =>x=0 hoặc x+3=0

=>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

\(a,\Leftrightarrow3\left(x+3\right)=0\Leftrightarrow x=-3\\ b,\Leftrightarrow\left(x^2-2\right)\left(6x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=2\\6x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\\x=-\dfrac{1}{6}\end{matrix}\right.\\ c,\Leftrightarrow\left(x-2013\right)\left(4x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2013\\x=\dfrac{1}{4}\end{matrix}\right.\\ d,\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x+1-1\right)=0\\ \Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Bài 2: 

a: =>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

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

\(\Leftrightarrow2x^2+5x-3-x^2+2x+3=0\)

\(\Leftrightarrow x^2+7x=0\Leftrightarrow x\left(x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)

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

\(\Leftrightarrow2x^2+5x-12-3x^2+12=0\)

\(\Leftrightarrow x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

tham khảo: https://hoc24.vn/cau-hoi/.2256230161739

a) ⇔ \(4x^2+4x-x-1=0\)

⇔ \(4x^2+3x-1=0\)

⇔ \(4x(x+1)-(x+1)=0\)

⇔ \((x+1)(4x-1)=0\)

⇒ \(\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy...

b) \(x^3-4x^2+4x=0\)

⇔ \(x^2(x-2)-2x(x-2)=0\)

⇔ \((x-2)(x^2-2x)=0\)

⇒ \(\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

Vậy...

c) \(x^2-3x+2=0\)

⇔ \(x(x-2)-(x-2)=0\)

⇔ \((x-1)(x-2)=0\)

⇒ \(\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Vậy...

a) (x-1)(x-2)=0

x-1=0 --> x = 1

x-2=0 --> x = 2

d) x^2(x + 1) + 27(x + 1)=0

(x+1)(x^2+27)=0

x+1=0 --> x = -1

x^2+27=0 (vô lí)

a) \(x\left(x-2\right)-x+2=0\)

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

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

TH1:x-1=0⇒x=1

TH2:x-2=0⇒x=2

a) 

TH1:x-1=0⇒x=1

TH2:x-2=0⇒x=2

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

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

\(\left(3-3x+1\right)\left(3-x-1\right)=0\)

TH1:3-3x+1=0⇒x\(=\dfrac{4}{3}\)

TH2:3-x-1=0⇒x=2

Mệnh đề A đúng

Do \(x^2-x+1=\left(x-1\right)^2+\dfrac{3}{4}>0;\forall x\in R\)