a) ( x 2  – 4x + 1)( x 2  – 2x + 3).

b) ( x 2  + 5x – 1)( x 2  + x – 1).

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

b: \(x^8+x^7+1\)

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

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

c: \(x^8+x^4+1\)

\(=\left(x^8+2x^4+1\right)-x^4\)

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

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

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

\(a)\; x^4+4 \\= x^4+4x^2+4-4x^2\\=(x^2+2)^2-4x^2\\=(x^2+2-2x)(x^2+2+2x)\)

a) \(40x^4-10x^2=10x^2\left(4x^2-1\right)=10x^2\left(2x-1\right)\left(2x+1\right)\)

b) \(16x^4-20x^2-y^2-5y=\left(4x^2-\dfrac{5}{2}\right)^2-\left(y-\dfrac{5}{2}\right)^2=\left(4x^2-\dfrac{5}{2}-y+\dfrac{5}{2}\right)\left(4x^2-\dfrac{5}{2}+y-\dfrac{5}{2}\right)=\left(4x^2-y\right)\left(4x^2+y-5\right)\)c)\(64a^2-9b^2-16a+1=\left(8a-1\right)^2-9b^2=\left(8a-1-3b\right)\left(8a-1+3b\right)\)d) \(5x^2+23x-10=5\left(x-\dfrac{2}{5}\right)\left(x+5\right)\)

a: \(40x^4-10x^2\)

\(=10x^2\left(4x^2-1\right)\)

\(=10x^2\cdot\left(2x-1\right)\left(2x+1\right)\)

b: \(16x^4-20x^2-y^2-5y\)

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

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

c: Ta có: \(64a^2-9b^2-16a+1\)

\(=\left(8a-1\right)^2-9b^2\)

\(=\left(8a-1-3b\right)\left(8a-1+3b\right)\)

d: Ta có: \(5x^2+23x-10\)

\(=5x^2+25x-2x-10\)

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

Phần h mk chưa nghĩ ra

Lời giải:

a.

$3x^2+xy-4y^2=(3x^2-3xy)+(4xy-4y^2)=3x(x-y)+4y(x-y)=(x-y)(3x+4y)$

b.

$x^8-5x^4+4=(x^8-x^4)-(4x^4-4)$

$=x^4(x^4-1)-4(x^4-1)=(x^4-1)(x^4-4)$

$=(x^2-1)(x^2+1)(x^2-2)(x^2+2)$

$=(x-1)(x+1)(x^2+1)(x-\sqrt{2})(x+\sqrt{2})(x^2+2)$

c.

$x^3+3x^2+3x-7=(x^3+3x^2+3x+1)-8$

$=(x+1)^3-2^3=(x+1-2)[(x+1)^2+2(x+1)+4]$

$=(x-1)(x^2+4x+7)$

a) \(3x^2+xy-4y^2=3x^2-3xy+4xy-4y^2\)

\(=3x(x-y)+4y(x-y)=(3x+4y)(x-y)\)

b)\(x^8-5x^4+4=x^8-x^4-4x^4+4\)

\(=x^2(x^4-1)-4(x^4-1)=(x^2-4)(x^4-1)\)

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

c)\(x^3+3x^2+3x-7=x^3+3x^2+3x+1-8\)

\(\left(x+1\right)^3-\sqrt{2}^3=\left(x+1-\sqrt[]{2}\right)\left(\left(x+1\right)^2+2\sqrt{2}x+2\right)\)

a: \(3x^2+xy-4y^2\)

\(=3x^2+4xy-3xy-4y^2\)

\(=x\left(3x+4y\right)-y\left(3x+4y\right)\)

\(=\left(3x+4y\right)\left(x-y\right)\)

b: \(x^8-5x^4+4\)

\(=x^8-x^4-4x^4+4\)

\(=x^4\left(x^4-1\right)-4\left(x^4-1\right)\)

\(=\left(x^4-4\right)\left(x^4-1\right)\)

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

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

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

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

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

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

2)  \(x^3-9x^2+6x+16\)

\(\left(x^3+1\right)-\left[\left(9x^2-6x+1\right)-16\right]\)

\(=\left(x^3+1\right)-\left[\left(3x-1\right)^2-16\right]=\left(x^3+1\right)-\left(3x-1+4\right)\left(3x-1-4\right)\)\(=\left(x^3+1\right)-3\left(3x-5\right)\left(x+1\right)\)\(=\left(x+1\right)\left[x^2-x+1-9x+15\right]=\left(x+1\right)\left(x^2-10x+16\right)\)

\(=\left(x+1\right)\left[x\left(x-2\right)-8\left(x-2\right)\right]\)\(\left(x+1\right)\left(x-2\right)\left(x-8\right)\)

3)   \(x^3-6x^2-x+30\)

\(=x^3-5x^2-x^2+5x-6x+30\)

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

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

4)  \(2x^3-x^2+5x+3=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)

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

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

5) \(27x^3-27x^2+18x-4=\left(27x^3-1\right)-\left(27x^2-18x+3\right)\)

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

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

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

gửi phần này trước còn lại làm sau !!! tk mk nka !!!

nhiều thế

6) \(\left(x+y\right)^2-\left(x+y\right)-12\)\(=\left(x+y\right)^2-2\cdot\frac{1}{2}\left(x+y\right)+\frac{1}{4}-\frac{49}{4}\)

\(=\left(x+y-\frac{1}{2}\right)^2-\left(\frac{7}{2}\right)^2\)\(=\left(x+y-\frac{1}{2}-\frac{7}{2}\right)\left(x+y-\frac{1}{2}+\frac{7}{2}\right)\)

\(=\left(x-4\right)\left(x+3\right)\)

7)   \(\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)          (NHÂN x + 2 vs x +  5  và  x + 3 vs x + 4 )

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

ĐẶT   \(x^2+7x+11=y\)   ta được :  

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

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

8)  \(4x^4-32x^2+1=4x^4+4x^2+1-36x^2\)

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

9) sai đề rùi bạn ơi ! đề đúng nè 

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

Ta thấy :  

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

Thay vào biểu thức bài cho ta được : 

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

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

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

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

bài ở trên câu 3 : kết luận là  \(\left(x-3\right)\left(x^2-x-6\right)\)bạn sửa lại giúp mk nka !!! Th@nk !!! Tk Mk vs  

\(a,x^2-4xa+4a^2-81y^2=\left(x-2a\right)^2-\left(9y\right)^2=\left(x-2a-9y\right)\left(x-2a+9y\right)\\ b,3x^2-8x+4=\left(3x^2-6x\right)-\left(2x-4\right)=3x\left(x-2\right)-2\left(x-2\right)=\left(x-2\right)\left(3x-2\right)\)

\(3,=\left(x-y\right)^3+\left(y-x+x-z\right)^3+\left(z-x\right)^3\\ =\left(x-y\right)^3+\left(y-x\right)^3+3\left(y-x\right)\left(x-z\right)\left(y-x+x-z\right)+\left(x-z\right)^3+\left(z-x\right)^3\\ =\left(x-y\right)^3-\left(x-y\right)^3+3\left(y-x\right)\left(x-z\right)\left(y-z\right)-\left(z-x\right)^3+\left(z-x\right)^3\\ =3\left(y-x\right)\left(x-z\right)\left(y-z\right)\)

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

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

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

c) \(x^2+7x+12=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)

a) (x+1)(x+4)

b) (x-1)(3x+7)

c) (x+3)(x+4)

a) x²+5x+4 = x(x+4)+(x+4) = (x+4)(x+1)

b) 3x²+4x-7 = 3x(x-1)+7(x-1) = (x-1)(3x+7)

c) x²+7x+12 = x(x+4)+3(x+4) = (x+3)(x+4)

Lời giải:
a.

$x^8+x^4+1=(x^4)^2+2x^4+1-x^4$
$=(x^4+1)^2-(x^2)^2=(x^4+1-x^2)(x^4+1+x^2)$

$=(x^4+1-x^2)[(x^2+1)^2-x^2]$

$=(x^4-x^2+1)(x^2+1-x)(x^2+1+x)$

b. 

$x^{12}-3x^6-1=(x^6-\frac{3}{2})^2-\frac{13}{4}$

$=(x^6-\frac{3}{2}-\frac{\sqrt{13}}{2})(x^6-\frac{3}{2}+\frac{\sqrt{13}}{2})$

c.

$3x^4+10x^2-25=(3x^4+15x^2)-(5x^2+25)$

$=3x^2(x^2+5)-5(x^2+5)=(x^2+5)(3x^2-5)$

$=(x^2+5)(\sqrt{3}x-\sqrt{5})(\sqrt{3}x+\sqrt{5})$

c.

$x^2-5y^2-y^4+2xy-9$

$=(x^2+2xy+y^2)-(y^4+6y^2+9)$
$=(x+y)^2-(y^2+3)^2$
$=(x+y+y^2+3)(x+y-y^2-3)$

\(a,x^8+x^4+1\\ =\left(x^8+2x^4+1\right)-x^4\\ =\left(x^4+1\right)^2-x^4\\ =\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)\\ b,x^{12}-3x^6-1\\ =\left(x^{12}-2x^6+1\right)-x^6-2\\ =\left(x^6-1\right)^2-x^6-2\\ =\left(x^6-x^3-1\right)\left(x^6+x^3-1\right)-2???\\ c,3x^4+10x^2-25\\ =4x^4-\left(x^4-10x^2+25\right)\\ =4x^4-\left(x^2-5\right)^2\\ =\left(2x^2-x^2+5\right)\left(2x^2+x^2-5\right)\\ =\left(x^2+5\right)\left(3x^2-5\right)\\ d,x^2-5y^2-y^4+2xy-9\\ =\left(x^2+2xy+y^2\right)-\left(y^4+6y^2+9\right)\\ =\left(x+y\right)^2-\left(y^2+3\right)^2\\ =\left(x+y+y^2+3\right)\left(x+y-y^2-3\right)\)

a) x⁸+x⁴+1 = (x⁴+1)²-x⁴ = (x⁴-x²+1)(x⁴+x²+1)

b) x¹²-3x⁶-1 = (x⁶-1)²-x⁶ = (x⁶-x³-1)(x⁶+x³-1)

c) 3x⁴+10x²-25 = 4x⁴-(x⁴-10x²+25) = 4x⁴- (x²-5)² = (x²+5)(3x²-5)

d) x²-5y²-y⁴+2xy-9 = (x+y)²-(y²+3)² = (x+y-y²-3)(x+y+y²+3)