a)

\(\Rightarrow\left[\left(x^3+3^3\right)+\left(x+3\right)\right].\left(x-9\right)=0\)

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

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

\(\Rightarrow\left[\left(x+3\right).\left(x^2-3x+10\right)\right].\left(x-9\right)=0\)

\(\Rightarrow\left(x+3\right)=0\) hoặc x-9=0 hoặc \(x^2-3x+10=0\)

1) x-9=0 => x=9

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

3) \(x^2-3x+10=0\Rightarrow x^2-2.x.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{31}{4}=0\)

\(\Rightarrow\left(x-\dfrac{3}{2}\right)^2+\dfrac{31}{4}=0\Rightarrow\left(x-\dfrac{3}{2}\right)^2=-\dfrac{31}{4}\left(ktm\right)\)

Vậy x=9 ; x=-3

tìm x biết

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

b.9x2-4-(3x-2)(4x-5)=0

c.x2(x+3)-x2-3x=0

a) \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\Leftrightarrow x^2-4x+4-\left(x^2-9\right)=6\)

\(\Leftrightarrow x^2-4x+4-x^2+9=6\Leftrightarrow4x=4+9-6\Leftrightarrow4x=7\)

\(\Leftrightarrow x=\dfrac{7}{4}\) vậy \(x=\dfrac{7}{4}\)

b) \(9x^2-4-\left(3x-2\right)\left(4x-5\right)=0\)

\(\Leftrightarrow9x^2-4-\left(12x^2-15x-8x+10\right)=0\)

\(\Leftrightarrow9x^2-4-12x^2+15x+8x-10=0\Leftrightarrow-3x^2+23x-14=0\)

\(\Leftrightarrow-3x^2+2x+21x-14=0\Leftrightarrow-x\left(3x-2\right)+7\left(3x-2\right)=0\)

\(\Leftrightarrow\left(7-x\right)\left(3x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}7-x=0\\3x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=7\\3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)

vậy \(x=7;x=\dfrac{2}{3}\)

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

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

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

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

vậy \(x=0;x=1;x=-3\)