`2)x^4+2x^3-x^2-2x+1=0`

`<=>x^4+2x^3+x^2-2x^2-2x+1=0`

`<=>(x^2+x)^2-2(x^2+x)+1=0`

`<=>(x^2+x-1)^2=0`

`<=>x^2+x-1=0`

`\Delta=1+4=5`

`=>x_{1,2}=(-1+-sqrt5)/2`

Vậy `S={(-1+sqrt5)/2,(-1+sqrt5)/2`

`3)x^4-4x^3-9x^2+8x+4=0`

`<=>x^4-x^3-3x^3+3x^2-12x^2+12x-4x+4=0`

`<=>(x-1)(x^3-3x^2-12x-4)=0`

`<=>(x-1)(x^3+2x^2-5x^2-10x-2x-4)=0`

`<=>(x-1)(x+2)(x^2-5x-10)=0`

`+)x=1`

`+)x=-2`

`+)x^2-5x-10=0`

`Delta=25+40=65`

`=>x_{12}=(5+sqrt{65})/2`

 2x^4-9x^3+14x^2-9x+2=0 
vế trái có tổng các hệ số (2-9+14-9+2)=0 nến có 1 nghiêm x=1 
nên phân tích đc nhân tử là (x-1) 
2x^4-9x^3+14x^2-9x+2=0 <=> (x-1)(2x^3-7x^2+7x-2)=0 
<=> x=1 và 2x^3-7x^2+7x-2=0 
PT: 2x^3-7x^2+7x-2=0 cũng có tổng các hệ số (2-7+7-2)=0 nên cũng có 1 nghiệm là 1 => vế trái có thể phân tích đc nhân tử (x-1) 
2x^3-7x^2+7x-2=0 <=> (x-1)(2x^2-5x+2)=0 
<=> x=1 và 2x^2-5x+2=0 
2x^2-5x+2=0 <=> x^2 - (5/2)x + 1 =0 
<=> (x-5/4)^2 - 9/16 = 0 
<=> (x-5/4)^2 - (3/4)^2 = 0

2x⁴-9x³+14x²-9x+2=0

<=> 2x⁴-2x³-7x³+7x²+7x²-7x-2x+2=0

<=> 2x³(x-1)-7x²(x-1)+7x(x-1)-2(x-1)=0

<=> (x-1)(2x³-7x²+7x-2)=0

<=> (x-1)[2x³-2x²-5x²+5x+2x-2]=0

<=> (x-1)[2x²(x-1)-5x(x-1)+2(x-1)]=0

<=> (x-1)²(2x²-5x+2)=0

<=> (x-1)²(2x²-4x-x+2)=0

<=> (x-1)²[(2x(x-2)-(x-2)]=0

<=> (x-1)²(x-2)(2x-1)=0

=> \(\hept{\begin{cases}\left(x-1\right)^2=0\\x-2=0\\2x-1=0\end{cases}}\) <=> \(\hept{\begin{cases}x_1=1\\x_2=2\\x_3=\frac{1}{2}\end{cases}}\)

Ta có : (x³ - 2x²) - 9x + 18 = 0

<=> x²(x - 2) - (9x - 18) = 0

<=> x²(x - 2) - 9(x - 2) = 0

=> (x² - 9) (x - 2) = 0

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=3;-3\\x=2\end{cases}}\)

\(2x^4-9x^3+14x^2-9x+2=0\)

\(\Leftrightarrow2x^4-4x^3-5x^3+10x^2+4x^2-8x-x+2=0\)

\(\Leftrightarrow2x^3\left(x-2\right)-5x^2\left(x-2\right)+4x\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x^3-5x^2+4x-1\right)=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(b.6x^4+25x^3+12x^2-25x+6=0\\\Leftrightarrow 6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\\\Leftrightarrow 6x^3\left(x+2\right)+13x^2\left(x+2\right)-14x\left(x+2\right)+3\left(x+2\right)=0\\\Leftrightarrow \left(6x^3+13x^2-14x+3\right)\left(x+2\right)=0\\ \Leftrightarrow\left(6x^3+18x^2-5x^2-15x+x+3\right)\left(x+2\right)=0\\\Leftrightarrow \left[6x^2\left(x+3\right)-5x\left(x+3\right)+\left(x+3\right)\right]\left(x+2\right)=0\\ \Leftrightarrow\left(6x^2-5x+1\right)\left(x+3\right)\left(x+2\right)=0\\ \Leftrightarrow\left(6x^2-3x-2x+1\right)\left(x+3\right)\left(x+2\right)=0\\\Leftrightarrow \left[3x\left(2x-1\right)-\left(2x-1\right)\right]\left(x+3\right)\left(x+2\right)=0\\\Leftrightarrow \left(3x-1\right)\left(2x-1\right)\left(x+3\right)\left(x+2\right)=0\)

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

Vậy tập nghiệm của phương trình trên là \(S=\left\{\frac{1}{3};\frac{1}{2};-3;-2\right\}\)

\(2x^4-9x^3+14x^2-9x+2=0\\\Leftrightarrow 2x^4-2x^3-7x^3+7x^2+7x^2-7x-2x+2=0\\\Leftrightarrow 2x^3\left(x-1\right)-7x^2\left(x-1\right)+7x\left(x-1\right)-2\left(x-1\right)=0\\\Leftrightarrow \left(2x^3-7x^2+7x-2\right)\left(x-1\right)=0\\\Leftrightarrow \left[2\left(x^3-1\right)-7x\left(x-1\right)\right]\left(x-1\right)=0\\\Leftrightarrow \left(x-1\right)^2\left[2\left(x^2+x+1\right)-7x\right]=0\\\Leftrightarrow \left(2x^2+2x+2-7x\right)\left(x-1\right)^2=0\\\Leftrightarrow \left(2x^2-5x+2\right)\left(x-1\right)^2=0\\\Leftrightarrow \left(2x^2-x-4x+2\right)\left(x-1\right)^2=0\\\Leftrightarrow \left[x\left(2x-1\right)-2\left(2x-1\right)\right]\left(x-1\right)^2=0\\\Leftrightarrow \left(x-2\right)\left(2x-1\right)\left(x-1\right)^2=0\)

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

Vậy tập nghiệm của phương trình trên là \(S=\left\{2;\frac{1}{2};1\right\}\)

Lời giải:

$2x^4-9x^3+14x^2-9x+2=0$

$\Leftrightarrow 2x^4-2x^3-7x^3+7x^2+7x^2-7x-2x+2=0$

$\Leftrightarrow 2x^3(x-1)-7x^2(x-1)+7x(x-1)-2(x-1)=0$

$\Leftrightarrow (x-1)(2x^3-7x^2+7x-2)=0$

$\Leftrightarrow (x-1)[2(x^3-1)-7x(x-1)]=0$

$\Leftrightarrow (x-1)(x-1)(2x^2+2x+2-7x)=0$

$\Leftrightarrow (x-1)^2(2x^2-5x+2)=0$

$\Leftrightarrow (x-1)^2(2x^2-4x-x+2)=0$

$\Leftrightarrow (x-1)^2[2x(x-2)-(x-2)]=0$

$\Leftrightarrow (x-1)^2(2x-1)(x-2)=0$

\(\Rightarrow \left[\begin{matrix} x=1\\ x=\frac{1}{2}\\ x=2\end{matrix}\right.\)