\(a,5x\left(x^2-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,3\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow3\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\ c,x^2-9x-10=0\\ \Leftrightarrow x^2+x-10x-10=0\\ \Leftrightarrow x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)

a, 5\(x\)(\(x^2\) - 9) = 0

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

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

Vậy \(x\) \(\in\) { -3; 0; 3}

b, 3.(\(x+3\)) - \(x^2\) - 3\(x\) = 0

    3.(\(x+3\)) - \(x\).( \(x\) + 3) = 0

    (\(x+3\))( 3 - \(x\)) = 0

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

Vậy \(x\) \(\in\){ -3; 3}

c, \(x^2\) - 9\(x\) - 10 = 0

   \(x^2\) + \(x\) - 10\(x\)  - 10 = 0

   \(x.\left(x+1\right)\) - 10.( \(x-1\)) = 0

        (\(x+1\))(\(x-10\)) = 0

         \(\left[{}\begin{matrix}x+1=0\\x-10=0\end{matrix}\right.\)

           \(\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)

Vậy \(x\) \(\in\){ -1; 10}

a) x² + 11x = 0

x(x + 11) = 0

x = 0 hoặc x + 11 = 0

*) x + 11 = 0

x = -11

Vậy x = -11; x = 0

b) (2x - 3)² - (x + 4)² = 0

(2x - 3 - x - 4)(2x - 3 + x + 4) = 0

(x - 7)(3x + 1) = 0

x - 7 = 0 hoặc 3x + 1 = 0

*) x - 7 = 0

x = 7

*) 3x + 1 = 0

3x = -1

x = -1/3

Vậy x = -1/3; x = 7

c) x² + 7x = 8

x² + 7x - 8 = 0

x² - x + 8x - 8 = 0

(x² - x) + (8x - 8) = 0

x(x - 1) + 8(x - 1) = 0

(x - 1)(x + 8) = 0

x - 1 = 0 hoặc x + 8 = 0

*) x - 1 = 0

x = 1

*) x + 8 = 0

x = -8

Vậy x = -8; x = 1

\(a,\\ \left(x+2\right)^2-x.\left(x-1\right)=10\\ \Leftrightarrow x^2+4x+4-x^2+x=10\\ \Leftrightarrow\left(x^2-x^2\right)+4x+x=10-4\\ \Leftrightarrow5x=6\\ \Leftrightarrow x=\dfrac{6}{5}\\ b,\\ x^3-6x^2+9x=0\\ \Leftrightarrow x.\left(x^2-6x+9\right)=0\\ \Leftrightarrow x.\left(x-3\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

a) 3x^3-12x=0

3x(x^2-4)=0

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

suy ra 3x=0       suy ra x=0

           x-2=0               x=2

           x+2=0              x= -2

b) (x-3)^2-(x-3)(3-x)^2=0

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

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

(x-3)^2(4-x)=0

suy ra x-3=0  suy ra x=3

          4-x=0             x=4

a) và b) đã nhé bạn

a) ta có : \(\left(x-1\right)^3+3x\left(x-4\right)+1=0\)

\(\Leftrightarrow x^3-3x^2+3x-1+3x^2-12x+1=0\)

\(\Leftrightarrow x^3-9x=0\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\\x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\) vậy \(x=0;x=\pm3\)

b) \(x^2-25=6x-9\Leftrightarrow x^2-6x-16=0\)

\(\Leftrightarrow x^2+2x-8x-16=0\Leftrightarrow x\left(x+2\right)-8\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

vậy \(x=8;x=-2\)

a )

\(\left(x-1\right)^3+3x\left(x-4\right)+1=0\)

\(\Leftrightarrow x^3-3x^2+3x-1+3x^2-12x+1=0\)

\(\Leftrightarrow x^3-9x=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm3\end{matrix}\right.\)

Vậy \(x\in\left\{0;\pm3\right\}\)

b )

\(x^2-25=6x-9\)

\(\Leftrightarrow x^2-6x=25-9\)

\(\Leftrightarrow x^2-6x=16\)

\(\Leftrightarrow x^2-6x-16=0\)

\(\Leftrightarrow x^2-2.x.3+9-25=0\)

\(\Leftrightarrow\left(x-3\right)^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

Vậy ...

yggucbsgfuyvfbsudy

????????

a. x mũ 2 - 2x + 1 = 25 

= x^2 + 2.x.1 + 1^2

= ( x + 1 ) ^2

ko bt có đúng ko nữa, mấy câu kia tui ko bt lm

Sos