\(\left|x^3+x\right|-\left|9x^2+9\right|=0\)

\(\Leftrightarrow\left|x\left(x^2+1\right)\right|-9\left|x^2+1\right|=0\)

\(\Leftrightarrow\left(\left|x\right|-9\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\left|x\right|=9\left(x^2+1\ge1>0\right)\Leftrightarrow x=\pm9\)

Vậy ... 

\(\left|x^3+x\right|-\left|9x^2+9\right|=0\)

\(TH1:\left\{{}\begin{matrix}\left|x^3+x\right|=0\\\left|9x^2+9\right|=0\end{matrix}\right.\)

\(\text{Vì }9x^2\ge0\)

\(\Rightarrow9x^2+9\ge9\)

\(TH2:\left|x^3+x\right|=\left|9x^2+9\right|\)

\(\Rightarrow\left[{}\begin{matrix}x^3+x=9x^2-9\\x^3+x=9x^2+9\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^3+x+9x^2+9=0\\x^3+x-9x^2-9=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x.\left(x^2+1\right)+9.\left(x^2+1\right)=0\\x.\left(x^2+1\right)-9.\left(x^2+1\right)=0\end{matrix}\right.\)

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

1, \(x^3+4x^2+4x=0\Leftrightarrow x\left(x^2+4x+4\right)=0\)

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

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

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

3, \(x^4-9x^2=0\Leftrightarrow x^2\left(x^2-9\right)=0\)

\(\Leftrightarrow x^2\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=0;\pm3\)

4, \(x^2-6x+9=81\Leftrightarrow\left(x-3\right)^2=9^2\)

\(\Leftrightarrow\left(x-3-9\right)\left(x-3+9\right)=0\Leftrightarrow\left(x-12\right)\left(x+6\right)=0\Leftrightarrow x=-6;x=12\)

5, em xem lại đề nhé

à lag tý @@

5, \(x^3+6x^2+9x-4x=0\Leftrightarrow x^3+6x^2+5x=0\)

\(\Leftrightarrow x\left(x^2+6x+5\right)=0\Leftrightarrow x\left(x^2+x+5x+5\right)=0\)

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

a)\(x^3+4x^2+4x=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x+2\right)^2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}}\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+3-2=0\\x+3+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-5\end{cases}}}\)

c)\(x^4-9x^2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}}\)

d)\(x^2-6x+9=81\)

\(\Leftrightarrow\left(x-3\right)^2=81\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=9\\x-3=-9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=12\\x=-6\end{cases}}}\)

e)\(x^3+6x^2+9x-4x=0\)

\(\Leftrightarrow x^3+6x^2+5x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2+5x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0;x=-5\\x=-1\end{cases}}}\)

#H

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

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) 5x(x2-9)=0

=> TH1 5x=0

      <=> x= 0

     TH2: 2x-9=0

       <=> 2x=9

       <=> x = \(\dfrac{9}{2}\)

b, 3(x+3) - x²- 3x = 0   

<=> 3x + 9 - x² -3x = 0

<=> - x² +9 = 0

<=> - x² = -9

 <=> x = 3

c, x² -9x -10 = 0

<=> x² -x + 10x -10 = 0

<=> x(x-1)+10(x-1)=0     

 <=> (x-1)(x+10)=0

=> TH1: x-1=0

       <=> x=1

      TH2: x +10=0

      <=>  x=-10

b) \(x^2-2x-3=0\)

\(D=b^2-4ac\)

\(\left(-2\right)^2-\left(4\left(1.3\right)\right)=16\)

\(x_{1,2}=\frac{-b-\sqrt{D}}{2a}=\frac{2-\sqrt{16}}{2}\)

\(x=1;-3\)

a)1/3
b)-1
c)3

\(\frac{x^3\left(-1\right)}{9}=-\frac{x^3}{9}\)

\(-\frac{x^3}{9}=0\Rightarrow x=0\)

( x - 2 ) ³ - ( x + 1 ) ³ + 9x ( x + 1 ) - 9 = 0
=> \(x^3-6x^2+12x-8-\left(x^3+3x^2+3x+1\right)+9x^2+9x-9=0\)
=> \(x^3-6x^2+12x-8-x^3-3x^2-3x-1+9x^2+9x-9=0\)
=> \(18x-18=0\)
=>         \(18x=0+18\)
=>         \(18x=18\)
=>             \(x=1\)

d) \(4x^2-9-x\left(2x-3\right)=0\)

\(\Leftrightarrow4x^2-9-2x^2+3x=0\)

\(\Leftrightarrow2x^2+3x-9=0\)

\(\Delta=3^2-4.2.\left(-9\right)=9+72=81\)

Vậy pt có 2 nghiệm phân biệt

\(x_1=\frac{-3+\sqrt{81}}{4}=\frac{-3}{2}\);\(x_1=\frac{-3-\sqrt{81}}{4}=-3\)

e) \(x^3+5x^2+9x=-45\)

\(\Leftrightarrow x^3+5x^2+9x+45=0\)

\(\Leftrightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)

\(\Leftrightarrow\left(x^2+9\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm3i\\x=-5\end{cases}}\)

f) \(x^3-6x^2-x+30=0\)

\(\Leftrightarrow\left(x^3-x^2-6x\right)-\left(5x^2-5x-30\right)=0\)

\(\Leftrightarrow x\left(x^2-x-6\right)-5\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-2x+3x-6\right)=0\)

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

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

\(\Leftrightarrow x\in\left\{5;-3;2\right\}\)

a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)

\(\Leftrightarrow24x+25=15\)

\(\Leftrightarrow24x=-10\)

hay \(x=-\dfrac{5}{12}\)

b) Ta có: \(2x^3-50x=0\)

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

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

c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)

\(\Leftrightarrow x^2+8x-9=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)

d) Ta có: \(x^3-x=0\)

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

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

e) Ta có: \(27x^3-27x^2+9x-1=1\)

\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)

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

\(\Leftrightarrow3x-1=1\)

\(\Leftrightarrow3x=2\)

hay \(x=\dfrac{2}{3}\)

\(\Leftrightarrow\left|x^3+x\right|=\left|9x^2+9\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}x^3+x=9x^2+9\\x^3+x=-9x^2+9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x^2+1\right)\left(x-9\right)=0\\\left(x^2+1\right)\left(x+9\right)=0\end{matrix}\right.\Leftrightarrow x=9\)