\(x^3-4x^2-9x+36=0\)

=> \(x^2\left(x-4\right)-9\left(x-4\right)=0\)

=> \(\left(x-4\right)\left(x^2-9\right)=0\)

=> \(\orbr{\begin{cases}x-4=0\\x^2-9=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=\pm3\end{cases}}\)

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

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

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

=> \(\left(x^2+x-12\right)\left(x^2-x-6\right)=0\)

=> \(\left(x^2-3x+4x-12\right)\left(x^2+2x-3x-6\right)=0\)

=> \(\left[x\left(x-3\right)+4\left(x-3\right)\right]\left[x\left(x+2\right)-3\left(x+2\right)\right]=0\)

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

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

=> \(\hept{\begin{cases}\left(x-3\right)^2=0\\x+4=0\\x+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\x=-4\\x=-2\end{cases}}\)

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

=> \(x^3-x-2x+2=0\)

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

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

=> x = 1

a) ( x - 5 )( x - 3 ) - ( x + 2 )( 2x - 1 ) + x² = 5

<=> x² - 8x + 15 - ( 2x² + 3x - 2 ) + x² = 5

<=> 2x² - 8x + 15 - 2x² - 3x + 2 = 5

<=> -11x + 17 = 5

<=> -11x = -12

<=> x = 12/11

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

<=> -2x² + 2x + x² + x - 3 = x + 4

<=> 3x - 3 = x + 4

<=> 3x - x = 4 + 3

<=> 2x = 7

<=> x = 7/2

Bài làm :

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

\(\Leftrightarrow x^2-3x-5x+15-\left(2x^2-x+4x-2\right)+x^2=5\)

\(\Leftrightarrow\left(x^2-2x^2+x^2\right)+\left(-3x-5x+x-4x\right)=5-2-15\)

\(\Leftrightarrow-11x=-12\)

\(\Leftrightarrow x=\frac{12}{11}\)

\(b,-2x\left(x-1\right)+\left(x-1\right)\left(2x+3\right)=x+4\)

\(\Leftrightarrow-2x^2+2x+2x^2+3x-2x-3-x=4\)

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

\(\Leftrightarrow2x=7\)

\(\Leftrightarrow x=\frac{2}{7}\)

Học tốt nhé

\(\cdot\left(x+1\right)^2\ge0\)

\(\Rightarrow x^2+2x+1>0\)

\(\Rightarrow2x^2+4x+2\ge0\)

 \(\Rightarrow\left(3x^2+3x+3\right)-\left(x^2-x+1\right)\ge0\)

\(\Rightarrow3\left(x^2+x+1\right)\ge x^2-x+1\)

\(\Rightarrow\)\(\frac{x^2+x+1}{x^2-x+1}\ge\frac{1}{3}\) (1)

\(\cdot\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow2x^2-4x+2\ge0\)

\(\Rightarrow\left(3x^2-3x+3\right)-\left(x^2+x+1\right)\ge0\)

\(\Rightarrow3\left(x^2-x+1\right)\ge x^2+x+1\)

\(\Rightarrow\frac{x^2+x+1}{x^2-x+1}\le3\)(2)

Từ(1),(2) => đpcm