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

a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)

\(=8x^5+2x^4-6x^3-14x^2\)

b: \(=2x^3-3x^2-5x+6x^2-9x-15\)

\(=2x^3+3x^2-14x-15\)

c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)

d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)

e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)

=2x^2-5x+1

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

b) \(x^4+6x^3+11x^2+6x+1=\left(x^4+3x^3+x^2\right)+\left(3x^3+9x^2+3x\right)+\left(x^2+3x+1\right)=x^2\left(x^2+3x+1\right)+3x\left(x^2+3x+1\right)+\left(x^2+3x+1\right)=\left(x^2+3x+1\right)^2\)

c) \(64x^4+1=\left[\left(8x^2\right)^2+16x^2+1\right]-16x^2=\left(8x^2+1\right)^2-\left(4x\right)^2=\left(8x^2-4x+1\right)\left(8x^2+4x+1\right)\)d) \(81x^4+4=\left[\left(9x^2\right)^2+36x^2+2^2\right]-36x^2=\left(9x^2+2\right)^2-\left(6x\right)^2=\left(9x^2-6x+2\right)\left(9x^2+6x+2\right)\)

Câu 1:

\(e,x^5-x^4-1=x^5-x^4+x^3-x^3+x^2-x^2+x-x-1\\ =\left(x^5-x^4-x^3\right)+\left(x^3-x^2-x\right)+\left(x^2-x-1\right)\\ =x^3\left(x^2-x-1\right)+x\left(x^2-x-1\right)+\left(x^2-x-1\right)\\ =\left(x^2-x-1\right)\left(x^3+x+1\right)\)

Câu 2:

\(a,\left(3x+ay+b\right)\left(x+cy+d\right)\\ =3x^2+3xcy+3xd+axy+acy^2+ayd+bx+bcy+bd\\ =3x^2+xy\left(3c+a\right)+x\left(b+3d\right)+y\left(ad+bc\right)+acy^2+bd\\ \Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}3c+a=-22\\b+3d=-4\end{matrix}\right.\\ad+bc=8\\\left\{{}\begin{matrix}ac=7\\bd=1\end{matrix}\right.\end{matrix}\right.\)

Xét \(bd=1\Leftrightarrow\left[{}\begin{matrix}b=1;d=1\\b=-1;d=-1\end{matrix}\right.\)

Với \(b=1;d=1\Leftrightarrow b+3d=1+3\cdot1=4\left(ktm\right)\)

Với \(b=-1;d=-1\Leftrightarrow b+3d=-1-3=-4\left(tm\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}3c+a=-22\\-a-c=8\\ac=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-1\\c=-7\end{matrix}\right.\)

Vậy \(3x^2-22xy-4x+8y+7y^2+1=\left(3x-y-1\right)\left(x-7y-1\right)\)

Cái chỗ ngoặc nhọn mà 5 dòng á a ko thấy trong cái phần công thức nên là ghi z chứ nó có 5 dòng đó nha

câu b tương tự, lười wa 😴

b mk thấy nó sai đề sao ý 

c) \(C=\left(x^2+x+4\right)^2+8x\left(x^2+x+4\right)+15x^2\)

\(=\left(x^2+x+4\right)^2+2.4x.\left(x^2+x+4\right)+16x^2-x^2\)

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

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

\(=\left(x^2+5x+4-x\right)\left(x^2+5x+4+x\right)=\left(x^2+4x+4\right)\left(x^2+6x+4\right)\)

1)

\(15x^3+29x^2-8x-12=(15x^3+30x^2)-(x^2+2x)-(6x+12)\)

\(=15x^2(x+2)-x(x+2)-6(x+2)\)

\(=(x+2)(15x^2-x-6)=(x+2)(15x^2-10x+9x-6)\)

\(=(x+2)[5x(3x-2)+3(3x-2)]\)

\(=(x+2)(3x-2)(5x+3)\)

2)

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

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

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

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

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

3.

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

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

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

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

4.

\(x^3+4x^2-31x-70=(x^3+2x^2)+(2x^2+4x)-(35x+70)\)

\(=x^2(x+2)+2x(x+2)-35(x+2)\)

\(=(x+2)(x^2+2x-35)\)

\(=(x+2)(x^2-5x+7x-35)\)

\(=(x+2)[x(x-5)+7(x-5)]\)

\(=(x+2)(x-5)(x+7)\)

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

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

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

\(D=x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)

\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)

\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)

\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)

B1 :

a) (2x - 1)²