2: Ta có: \(x^4-4x^3-9x^2+8x+4=0\)

\(\Leftrightarrow x^4-x^3-3x^3+3x^2-12x^2+12x-4x+4=0\)

\(\Leftrightarrow x^3\left(x-1\right)-3x^2\left(x-1\right)-12x\left(x-1\right)-4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-3x^2-12x-4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2-5x^2-10x-2x-4\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x^2-5x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=\dfrac{5-\sqrt{33}}{2}\\x=\dfrac{5+\sqrt{33}}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{1;-2;\dfrac{5-\sqrt{33}}{2};\dfrac{5+\sqrt{33}}{2}\right\}\)

1: Ta có: \(x^4+5x^3+10x^2+15x+9=0\)

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

\(\Leftrightarrow x^3\left(x+1\right)+4x^2\left(x+1\right)+6x\left(x+1\right)+9\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3+4x^2+6x+9\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left[x^3+3x^2+x^2+6x+9\right]=0\)

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

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

mà \(x^2+x+3>0\forall x\)

nên (x+1)(x+3)=0

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

Vậy: S={-1;-3}

a) Đa thức thương  x 2  – 6x + 9.

b) Đa thức thương 2 x 2  – 5.

c) Đa thức thương  x 2  + 4x + 3 và đa thức dư -12.

d) Đa thức x + 5 và đa thức dư x – 4.

`1)x^4 -10x^3 +26x^2 -10x+1=0`
`x=0=>VT=1=>x=0(l)`
Chia 2 vế cho `x^2>0` ta có
`x^2-10x+26-10/x+1/x^2=0`
`=>x^2+1/x^2+26-10(x+1/x)=0`
`=>(x+1/x)^2-10(x+1/x)+24=0`
Đặt `a=x+1/x`
`pt<=>a^2-10a+24=0`
`<=>` $\left[ \begin{array}{l}a=4\\a=6\end{array} \right.$
`a=4<=>x+1/x=4<=>x^2-4x+1=0<=>` $\left[ \begin{array}{l}x=\sqrt3+2\\x=-\sqrt3+2\end{array} \right.$
`a=6<=>x+1/x=6<=>x^2-6x+1=0<=>` $\left[ \begin{array}{l}x=\sqrt8+3\\x=-\sqrt8+3\end{array} \right.$
Vậy `S={\sqrt3+2,-\sqrt3+2,\sqrt8+3,-\sqrt8+3}`

2)Do hệ số chẵn bằng=hệ số lẻ
`=>x=-1`
`pt<=>x^4+x^3+4x^3+4x^2+6x^2+6x+9x+9=0`
`<=>(x+1)(x^3+4x^2+6x+9)=0`
`<=>(x+1)(x^3+3x^2+x^2+6x+9)=0`
`<=>(x+1)[x^2(x+3)+(x+3)^2]=0`
`<=>(x+1)(x+3)(x^2+x+3)=0`
Do `x^2+x+3=(x+1/2)^2+11/4>0`
`=>` $\left[ \begin{array}{l}x=-3\\x=-1\end{array} \right.$
Vậy `S={-1,-3}`

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

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

\(\Leftrightarrow x\left(x-4\right)-\left(x-4\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=1\end{cases}}}\)

\(x^2+15x+56=0\)

\(\Leftrightarrow x^2+7x+8x+56=0\)

\(\Leftrightarrow x\left(x+7\right)+8\left(x+7\right)=0\)

\(\Leftrightarrow\left(x+7\right)\left(x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+7=0\\x+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-7\\x=-8\end{cases}}}\)

=.= hk tốt!!

1) x² - 5x + 4 = 0

<=> (x - 1)(x - 4) = 0

<=> x - 1 = 0 hoặc x - 4 = 0

       x = 0 + 1         x = 0 + 4

       x = 1               x = 4

=> x = 1 hoặc x = 4

2) x² + 15x + 56 = 0

<=> (x + 7)(x + 8) = 0

<=> x + 7 = 0 hoặc x + 8 = 0

       x = 0 - 7           x = 0 - 8

        x = -7              x = -8

=> x = -7 hoặc x = -8

       x² - 5x + 4 = 0 

=> ( x - 1 ) ( x - 4 ) = 0

=>  x- 1 = 0                    x - 4 = 0

      x     = 1                    x      = 4

                                                              ~~Hok tốt~~

CM: 5x^2 +15x+20>0

Ta có: 5x^2 +15x +20

= 5( x^2 + 3x +4) 

=5[(x^2 + 2.x.3/2 +9/4) -9/4 +4 ]

=5(x+3/2)^2 -7/4

Vì (x+3/2)^2 >0 với mọi x

=>5(x+3/2)^2 >0 với mọi x

=> 5(x+3/2)^2 - 7/4 >0 với mọi x

a: =>6(2x-5)-3x(2x-5)=0

=>(2x-5)(6-3x)=0

=>x=5/2 hoặc x=2

b: \(\Leftrightarrow x^2+5x+4+x^2=x^2+4x+4\)

=>x²+x=0

=>x(x+1)=0

=>x=0(loại) hoặc x=-1(nhận)

a)4.(x+3)-(2x-12)=x-(-11+4)

4x+12-2x+12=x+11-4

2x+24=x+7

2x-x=-24+7

x=-17

b)-(4x-13)+(5x-4)=-3-(-15+7)

-4x+13+5x-4=-3+15-7

(-4x+5x)+13-4=12-7

x+9=5

x=-4

c)(5x-3)-(-2x+4)=6x-12

5x-3+2x-4=6x-12

5x+2x-6x=3+4-12

x=-5

d)(15x+20)-(9x-3)=5x-(-12)

15x+20-9x+3=5x+12

15x-9x-5x=-20-3+12

x=-11

e,(7x+14)+(3x-8)=-(-9x+3)

7x+14+3x-8=9x-3

7x+3x-9x=-14+8-3

x=-9