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

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

b)\(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\Rightarrow\left[3\left(2x+1\right)+2\left(x+1\right)\right]\left[3\left(2x+1\right)-2\left(x+1\right)\right]=0\)

\(\Rightarrow\left[8x+5\right]\left[4x+1\right]=0\Rightarrow\left[{}\begin{matrix}8x+5=0\\4x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)

c)\(x^3-6x^2+9x=0\Rightarrow x\left(x^2-6x+9\right)=0\Rightarrow x\left(x-3\right)^2=0\)

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

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

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

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

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

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

Do \(\left(x+1\right)^2+1>0\)

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

\(a,\Leftrightarrow x^3-8-x^3-2x=12\Leftrightarrow-2x=20\Leftrightarrow x=-10\\ b,\Leftrightarrow x^2-6x+9-x^2+4=16\Leftrightarrow=-6x=3\Leftrightarrow x=-\dfrac{1}{2}\\ c,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-6\right)+9\left(x-6\right)=0\\ \Leftrightarrow\left(x^2+9\right)\left(x-6\right)=0\\ \Leftrightarrow x=6\left(x^2+9>0\right)\)

f: Ta có: \(x\left(2x-9\right)-4x+18=0\)

\(\Leftrightarrow\left(2x-9\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=2\end{matrix}\right.\)

g: Ta có: \(4x\left(x-1000\right)-x+1000=0\)

\(\Leftrightarrow\left(x-1000\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1000\\x=\dfrac{1}{4}\end{matrix}\right.\)

f. x(2x - 9) - 4x + 18 = 0

<=> x(2x - 9) - 2(2x - 9) = 0

<=> (x - 2)(2x - 9) = 0

<=> \(\left[{}\begin{matrix}x-2=0\\2x-9=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=2\\x=\dfrac{9}{2}\end{matrix}\right.\)

g. 4x(x - 1000) - x + 1000 = 0

<=> 4x(x - 1000) - (x - 1000) = 0

<=> (4x - 1)(x - 1000) = 0

<=> \(\left[{}\begin{matrix}4x-1=0\\x-1000=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=1000\end{matrix}\right.\)

h. 2x(x - 4) - 6x²(-x + 4) = 0

<=> 2x(x - 4) + 6x²(x - 4) = 0

<=> (2x + 6x²)(x - 4) = 0

<=> 2x(1 + 3x)(x - 4) = 0

<=> \(\left[{}\begin{matrix}2x=0\\1+3x=0\\x-4=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{3}\\x=4\end{matrix}\right.\)

i. 2x(x - 3) + x² - 9 = 0

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

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

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

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

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

j. 9x - 6x² + x³ = 0

<=> x(9 - 6x + x²) = 0

<=> x(3 - x)² = 0

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

<=> \(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

 4x(x – 1000) – x + 1000 = 0

(4x-1)(x-1000)                 =0

⇔x=1/4 hoặc  1000

a) x = 1; x = - 1 3                 b) x = 2.

c) x = 3; x = -2.                 d) x = -3; x = 0; x = 2.

a: Ta có: \(\left(3x-2\right)\left(2x-1\right)-\left(6x^2-3x\right)=0\)

\(\Leftrightarrow2x-1=0\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x^3-\left(x+1\right)\left(x^2-x+1\right)=x\)

\(\Leftrightarrow x^3-x^3-1=x\)

hay x=-1

c: Ta có: \(56x^4+7x=0\)

\(\Leftrightarrow7x\left(8x^3+1\right)=0\)

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

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

d: Ta có: \(x^2-5x-24=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)

a: (1-2x)^3-(1+2x)^3

\(=1^3-3\cdot1^2\cdot2x+3\cdot1\cdot\left(2x\right)^2-8x^3-8x^3-12x^2-6x-1\)

\(=1-6x+12x^2-8x^3-8x^3-12x^2-6x-1\)

\(=-16x^3-12x\)

b: \(=x^3-6x^2+12x-8-x^3-x^2+8\)

\(=-7x^2+12x\)

c: \(=x^3+8-12x+6x^2-x^3+6x^2+12x\)

\(=12x^2+8\)

c: Ta có: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)

\(\Leftrightarrow3x^2+26x=0\)

\(\Leftrightarrow x\left(3x+26\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)

\(a,\Leftrightarrow x^2+8x+16-x^3-12x^2=16\\ \Leftrightarrow x^3+11x^2-8x=0\\ \Leftrightarrow x\left(x^2+11x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+11x-8=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=121+32=153\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11-3\sqrt{17}}{2}\\x=\dfrac{-11+3\sqrt{17}}{2}\end{matrix}\right.\\ S=\left\{0;\dfrac{-11-3\sqrt{17}}{2};\dfrac{-11+3\sqrt{17}}{2}\right\}\)

\(c,\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\\ \Leftrightarrow3x^2+26x=0\\ \Leftrightarrow x\left(3x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\\ d,\Leftrightarrow x^3-6x^2+12x-8-x^3-125-6x^2=11\\ \Leftrightarrow-12x^2+12x-144=0\\ \Leftrightarrow x^2-x+12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)