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) Ta có: \(8x\left(2x-3\right)-4x\left(4x+3\right)=72\)

\(\Leftrightarrow16x^2-24x-16x^2-12x=72\)

\(\Leftrightarrow-36x=72\)

hay x=-2

b) Ta có: \(\left(x+2\right)\left(x+4\right)-x\left(x+2\right)=104\)

\(\Leftrightarrow x^2+6x+8-x^2-2x=104\)

\(\Leftrightarrow4x=96\)

hay x=24

c) Ta có: \(\left(x-1\right)\left(x+4\right)-x\left(x-1\right)=308\)

\(\Leftrightarrow x^2+3x-4-x^2+x=308\)

\(\Leftrightarrow4x=312\)

hay x=78

d) Ta có: \(15x\left(2x-3\right)-\left(5x+2\right)\left(6x-5\right)=-22\)

\(\Leftrightarrow30x^2-45x-30x^2+25x-12x+10=-22\)

\(\Leftrightarrow-32x=-32\)

hay x=1

a) \(x^3+4x^2-29x+24=x^3-x^2+5x^2-5x-24x+24\)

\(=x^2\left(x-1\right)+5x\left(x-1\right)-24\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+5x-24\right)\)

\(=\left(x-1\right)\left(x^2+8x-3x-24\right)\)

\(=\left(x-1\right)\left[x\left(x+8\right)-3\left(x+8\right)\right]\)

\(=\left(x-1\right)\left(x+8\right)\left(x-3\right)\)

b) \(x^4+6x^3+7x^2-6x+1\)

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

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

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

c) \(\left(x^2-x+2\right)^2+\left(x-2\right)^2=x^4-2x^3+6x^2-8x+8\)

\(=\left(x^4-2x^3+2x^2\right)+\left(4x^2-8x+8\right)\)

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

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

d) Phức tạp mà dài quá :v

\(6x^5+15x^4+20x^3+15x^2+6x+1\)

\(=6x^5+3x^4+12x^4+6x^3+14x^3+7x^2+8x^2+4x+2x+1\)

\(=3x^4\left(2x+1\right)+6x^3\left(2x+1\right)+7x^2\left(2x+1\right)+4x\left(2x+1\right)+\left(2x+1\right)\)

\(=\left(2x+1\right)\left(3x^4+6x^3+7x^2+4x+1\right)\)

\(=\left(2x+1\right)\left[\left(3x^4+3x^3+x^2\right)+\left(3x^3+3x^2+x\right)+\left(3x^2+3x+1\right)\right]\)

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

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

e)

- Câu này có thể áp dụng định lý: nếu tổng các hệ số biến bậc chẵn và tổng các hệ số biến bậc lẻ bằng nhau thì đa thức có nhân tử x + 1.

- Nhận thấy: 1 + 4 + 4 + 1 = 3 + 4 + 3

\(x^6+3x^5+4x^4+4x^3+4x^2+3x+1\)

\(=(x^6+x^5)+(2x^5+2x^4)+(2x^4+2x^3)+(2x^3+2x^2)+(2x^2+2x)+(x+1)\)

\(=x^5(x+1)+2x^4(x+1)+2x^3(x+1)+2x^2(x+1)+2x(x+1)+(x+1)\)

\(=(x+1)(x^5+2x^4+2x^3+2x^2+2x+1)\)

Tiếp tục phân tích bằng cách trên vì 1 + 2 + 2 = 2 + 2 +1

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

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

a) Gọi CT ghi hóa trị của NH₃ là \(N^xH^I_3\) (x: nguyên, dương)

Theo quy tắc hóa trị, ta có:

\(x.1=I.3\\ =>x=\dfrac{1.I}{3}=III\)

Vậy: Hóa trị của N có hóa trị III trong hợp chất NH₃

b) Gọi CT kèm hóa trị của Zn(OH)₂ là \(Zn^x\left(OH\right)^y_2\) (x,y: nguyên, dương).

Theo quy tắc hóa trị, ta có:

\(x.1=y.2\\ =>\dfrac{x}{y}=\dfrac{2}{1}=\dfrac{II}{I}\)

=> x=II

y=I

=> Hóa trị của Zn là II trong hợp chất trên

1 

(3x-2)(4x+5)=0

⇔ 3x-2=0  -> x= 2/3      

 ⇔ 4x-5=0     x= 5/4

Vậy tập nghiệm S = { 2/3; 5/4}

2,    (4x+2)(\(X^2\)+3)=0

⇔ 4x+2=0         ->   x= -1/2    

     \(x^2\)+3=0         -> x= \(\sqrt{3}\); -\(\sqrt{3}\)

Vaayj tập nghiệm S= { -1/2; \(\sqrt{3}\);-\(\sqrt{3}\)}

3)

    (2x+7)(x-3)(5x-1)=0

⇔ 2x+7=0          ->   x= -7/2

     x-3   =0         ->     x =  3

    5x-1  =0         ->    x=  1/5

Vậy tập nghiệm S={ -7/2; 3; 1/5}

`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}`

b: \(\Leftrightarrow\dfrac{x-2}{A}=\dfrac{\left(5x-1\right)\left(x-2\right)}{x^2\left(5x-1\right)+3\left(5x-1\right)}=\dfrac{x-2}{x^2+3}\)

hay \(A=x^2+3\)

\(1,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 2,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 3,\left(x+1\right)^2+2\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x+1+2\right)=0\\ \Rightarrow\left(x+1\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

x²+4x+4=0
(x+2)²=0
x+2=0
x=+-2
câu 1 giống câu 2
(x+1)²+2(x+1)=0
(x+1+2)(x+1)=0
Th1: x+3=0           Th2: x+1=0
            x=-3                      x=-1
vậy ...