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giúp mình vớiiii

c)  \(x^3-9x^2+6x+16=x^3-8x^2-x^2+8x-2x+16\)

\(=x^2\left(x-8\right)-x\left(x-8\right)-2\left(x-8\right)=\left(x-8\right)\left(x^2-x-2\right)=\left(x-8\right)\left(x-2\right)\left(x+1\right)\)

d) \(2x^3+3x^2+3x+1=\left(2x+1\right)\left(x^2+x+1\right)\)

e)  \(2x^3-5x^2+5x-3=\left(2x-3\right)\left(x^2-x+1\right)\)

b) \(x^3-6x^2+9x=0\)

\(\Leftrightarrow x.\left(x^2-6x+9\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

Vậy \(x=0\)hoặc \(x=3\)

a. ( x - 1 )³ + 1 + 3x ( x - 4 ) = 0

<=> x³ - 3x² + 3x - 1 + 1 + 3x² - 12x = 0

<=> x³ - 9x = 0

<=> x ( x² - 9 ) = 0

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

b. x³ - 6x² + 9x = 0

<=> x ( x² - 6x + 9 ) = 0

<=> x ( x - 3 )² = 0

<=> \(\orbr{\begin{cases}x=0\\\left(x-3\right)^2=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

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

\(\Leftrightarrow9x^2-1+3x^2+6x-x-2=0\)

\(\Leftrightarrow9x^2+3x^2+6x-x=0+1+2\)

\(\Leftrightarrow12x^2+5x=3\)

\(\Leftrightarrow12x^2+5x-3=0\)

\(\Leftrightarrow12x^2-4x+9x-3=0\)

\(\Leftrightarrow4x\left(3x-1\right)+3\left(3x-1\right)\)

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

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

Vậy tập nghiệm phương trình là S = \(\left\{\dfrac{-3}{4};\dfrac{1}{3}\right\}\)

( 3x - 1 )( x + 3 ) + 9x² - 1 = 0

<=> 3x² + 9x - x - 3 + 9x² - 1 = 0

<=> 12x² + 8x - 4 = 0

<=> 4( 3x² + 2x - 1 ) = 0

<=> 3x² + 2x - 1 = 0 

<=> 3x² + 3x - x - 1 = 0

<=> ( 3x² + 3x ) - ( x + 1 ) = 0

<=> 3x( x + 1 ) - 1( x + 1 ) = 0

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

<=> \(\orbr{\begin{cases}3x-1=0\\x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=-1\end{cases}}\)

Vậy S = { 1/3 ; -1 }

\(\frac{x+1}{3}>\frac{3x-2}{5}\)

\(\Leftrightarrow\frac{5\left(x+1\right)}{15}>\frac{3\left(3x-2\right)}{15}\)

\(\Leftrightarrow5x+5>9x-6\)

\(\Leftrightarrow5x-9x>-6-5\)

\(\Leftrightarrow-4x>-11\)

\(\Leftrightarrow x< \frac{11}{4}\)

Bài làm:

a) \(\left(3x-1\right)\left(x+3\right)+9x^2-1=0\)

\(\Leftrightarrow3x^2+8x-3+9x^2-1=0\)

\(\Leftrightarrow12x^2+8x-4=0\)

\(\Leftrightarrow3x^2+2x-1=0\)

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

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)

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

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

Vậy tập nghiệm của PT \(S=\left\{-1;\frac{1}{3}\right\}\)

b) \(\frac{x+1}{3}>\frac{3x-2}{5}\Leftrightarrow\frac{5\left(x+1\right)}{15}>\frac{3\left(3x-2\right)}{15}\)

\(\Rightarrow5x+5>9x-6\)

\(\Leftrightarrow4x< 11\)

\(\Rightarrow x< \frac{11}{4}\)

Có : 

b) (x - 8)(x + 8) = (x - 4)(x² + 4x + 16)

  x² - 8² = x³ - 4³

x² - 2^6 - x³ + 2⁶  = 0

x² . ( x - 1 ) = 0

x²  = 0 hoặc x-1 = 0

x= 0 hoặc x = 1

 Vâỵ....

b)=x^3+x^2+4x^2+4x+4x+4=x(x+1)+4x(x+1)+4(x+1)=(x+1)(x+4x+4)

Thiên Ân ơi, bạn giải giúp mình câu c được kh