a: \(x\left(x-3\right)+2x-6=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

b: \(\left(x+1\right)^2-4\left(x+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)

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

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

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

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

e: ta có: \(4x^2+4x-6=2\)

\(\Leftrightarrow4x^2+4x-8=0\)

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

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

f: Ta có: \(2x^2+7x+3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

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

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

\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

a: Ta có: \(\left(3x+5\right)^2-4x^2=0\)

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

\(\Leftrightarrow\left(x+1\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)

\(a,\Leftrightarrow x^2-16-x^2-2x=0\\ \Leftrightarrow2x=-16\Leftrightarrow x=-8\\ b,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ c,\Leftrightarrow6x\left(1-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\\ d,\Leftrightarrow12x-56x^2+x^2-16=0\\ \Leftrightarrow55x^2-12x+16=0\\ \Delta=144-4\cdot55\cdot16< 0\\ \Leftrightarrow x\in\varnothing\)

a) Ta có: \(7x^2-28=0\)

\(\Leftrightarrow7\left(x^2-4\right)=0\)

\(\Leftrightarrow7\left(x-2\right)\left(x+2\right)=0\)

mà 7>0

nên (x-2)(x+2)=0

hay \(\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{2;-2\right\}\)

b) Ta có: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)

\(\Leftrightarrow\dfrac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)

mà \(\dfrac{2}{3}>0\)

nên x(x-2)(x+2)=0

hay \(\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-2;2\right\}\)

c) Ta có: \(2x\left(3x-5\right)-\left(5-3x\right)=0\)

\(\Leftrightarrow2x\left(3x-5\right)+\left(3x-5\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}3x-5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=5\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{5}{3};-\dfrac{1}{2}\right\}\)

d) Ta có: \(\left(2x-1\right)^2-25=0\)

\(\Leftrightarrow\left(2x-1-5\right)\left(2x-1+5\right)=0\)

\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{3;-2\right\}\)

a,7x² - 28 = 0

=> 7x² = 28 => x² = 4 => x = 2

b,2/3x(x² - 4) = 0

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

=> x ∈ {0 ; 2 ; -2}

c,2x(3x - 5) - (5 - 3x) = 0

= 2x(3x - 5) + (3x - 5)

= (3x - 5)(2x + 1) = 0

=> x ∈ { 5/3 ; -1/2}

d, (2x - 1)² - 25 = 0

=> (2x - 4)(2x - 6) = 0

=> x ∈ {2 ;3}

a,7x² - 28 = 0

=> 7x² = 28 => x² = 4 => x = 2

b,2/3x(x² - 4) = 0

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

=> x ∈ {0 ; 2 ; -2}

c,2x(3x - 5) - (5 - 3x) = 0

= 2x(3x - 5) + (3x - 5)

= (3x - 5)(2x + 1) = 0

=> x ∈ { 5/3 ; -1/2}

d, (2x - 1)² - 25 = 0

=> (2x - 4)(2x - 6) = 0

=> x ∈ {2 ;3}