a) x¹² + 4 = x¹² + 4x⁶ + 4 - 4x⁶ = (x⁶ + 2)² - (2x³)² 

= (x⁶ - 2x³ + 2)(x⁶ + 2x³ + 2)

b) 4x⁸ + 1 = 4x⁸ + 4x⁴  + 1 - 4x⁴ = (2x⁴ + 1)² - (2x²)² 

= (2x⁴ + 2x² + 1)(2x⁴ - 2x²  + 1)

c) x⁷ + x⁵ - 1 = x⁷ - x + x⁵ + x² - (x² - x  + 1) = x(x⁶ - 1) + x²(x³ + 1) - (x² - x + 1)

= x(x³ - 1)(x³ + 1) + x²(x + 1)(x² - x + 1) - (x2 - x + 1)

= (x⁴ - x)(x + 1)(x² - x + 1) + (x³ + x²)(x² - x + 1) - (x² - x + 1)

= (x⁵ + x⁴ - x² - x + x³ + x² - 1)(x² -x + 1)

= (x⁵ + x⁴ + x³ - x - 1)(x² - x + 1)

d) x⁷ + x⁵ + 1 = x⁷ - x + x⁵ - x² + (x² + x + 1)

= x(x³ - 1)((x³ + 1) + x²(x³ - 1) + (x² + x + 1)

= (x⁴ + x)(x  - 1)(x² + x + 1) + x²(x - 1)((x² + x + 1) + (x² + x + 1)

= (x² + x + 1)(x⁵ - x⁴ + x² - x + x³ - x² + 1)

= (x² + x + 1)(x⁵ - x⁴ + x³ - x + 1)

a) \(x^5-x^4-1\)

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

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

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

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

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

b) \(x^8+x^7+1\)

\(=\left(x^8-x^2\right)+\left(x^7-x\right)+\left(x^2+x+1\right)\)

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

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

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

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

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

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

a) \(x^5-x^4-1=x^5+x^2-x^4-x^2-1\)

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

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

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

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

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

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

b) \(x^8+x^7+1=x^8+x^7+x^6-x^6+1\)

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

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

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

Mong cô Chuy cho e thêm 1 Gp nựa nha cô '-'

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

a/ \(x^4+16\)

\(=x^4+4x^2+16-4x^2\)

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

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

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

b/ \(64x^4+y^4\)

\(=64x^4+y^4+16x^2y^2-16x^2y^2\)

\(=\left(64x^4+y^4+16x^2y^2\right)-16x^2y^2\)

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

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

ủa phần a mình phân tích rồi mà bạn hu hu

a) x⁷+ x² + 1

=x⁷-x+x²+x+1

=x.(x⁶-1)+(x²+x+1)

=x.(x³-1)(x³+1)+(x²+x+1)

=x.(x-1)(x²+x+1)(x³+1)+(x²+x+1)

=(x²+x+1)[x.(x-1)(x³+1)+1]

=(x²+x+1)(x⁵+x²-x⁴-x+1)

b) x⁵ + x⁴ + 1

=x⁵+x⁴+x³+x²+x+1-x³-x²-x

=x³.(x²+x+1)+(x²+x+1)-x.(x²+x+1)

=(x²+x+1)(x³+1-x)

\(x^8+x+1\)

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

\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)

\(x^5+x^4+1\)

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

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

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

\(\left(x+1\right)\left(x+3\right)\left(x+4\right)\left(x+6\right)-7\)

\(=\left\{\left(x+1\right)\left(x+6\right)\right\}.\left\{\left(x+3\right)\left(x+4\right)\right\}-7\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+12\right)-7\) \(\left(1\right)\)

đặt \(x^2+7x+9=a\)

<=> \(\left(1\right)=\left(a-3\right)\left(a+3\right)-7\)

             \(=a^2-16\)

               \(=\left(a-4\right)\left(a+4\right)\)

hay\(\left(1\right)=\) \(\left(x^2+7x+9-4\right)\left(x^2+7x+9+4\right)\)

               \(=\left(x^2+7x+5\right)\left(x^2+7x+13\right)\)

những câu còn lại cũng nhóm đầu với cuối , hai cái giữa với nhau , xong làm tương tự câu trên

học tốt

a) (x + 1)(x + 3)(x + 4)(x + 6) - 7

= (x + 1)(x + 6) (x + 3)(x + 4) - 7

= (x² + 7x + 6)(x + 7x + 12) - 7

Đặt t = x² + 7x + 6

Ta có : t(t + 6) - 7 

= t² + 6t - 7

= t² + 6t + 9 - 16 

= (t + 3) - 16

= (t + 3 - 4)(t + 3 + 4)

= (t - 1)(t + 7)

Nên : 

Pt = (x² + 7x + 6 - 1)(x² + 7x + 6 + 7)

=   (x² + 7x + 5)(x² + 7x + 13)

Phân tích đa thức thành nhân tử:

a) (x+1)(x+3)(x+4)(x+6)-7

b)(x+2)(x+3)(x+5)(x+6)-10

c) x(2x+1)(2x+3)(4x+8)-18

Làm : 

a) (x + 1)(x + 3)(x + 4)(x + 6) - 7

= (x + 1)(x + 6) (x + 3)(x + 4) - 7

= (x² + 7x + 6)(x + 7x + 12) - 7

Đặt t = x² + 7x + 6

Ta có : t(t + 6) - 7 

= t² + 6t - 7

= t² + 6t + 9 - 16 

= (t + 3) - 16

= (t + 3 - 4)(t + 3 + 4)

= (t - 1)(t + 7)

Nên : 

Pt = (x² + 7x + 6 - 1)(x² + 7x + 6 + 7)

=   (x² + 7x + 5)(x² + 7x + 13)