a: =(x^2+x)^2+3(x^2+x)-10

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

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

b: (x^2+2x)^2-2(x^2+2x)-3

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

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

c: =(x^2+x)^2+4(x^2+x)-12

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

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

d: =(x^2+10x+16)(x^2+10x+24)+16

=(x^2+10x)^2+40(x^2+10x)+400

=(x^2+10x+20)^2

giúp mik gấp vs ạ đag cần trong đêm nay

a) Ta có: \(x-2y+x^2-4y^2\)

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

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

b) Ta có: \(x^2+2xy+y^2-4x^2y^2\)

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

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

c) Ta có: \(x^6-x^4+2x^3+2x^2\)

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

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

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

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

d) Ta có: \(x^3+3x^2+3x+1-8y^3\)

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

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

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

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

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

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

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

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

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

d, \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-\left(2y\right)^3=\left(x+1-2y\right)\left(x+1+2y\right)\)

a) Ta có: \(x-2y+x^2-4y^2\)

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

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

b: Ta có: \(x^2-4x^2y^2+y^2+2xy\)

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

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

a: 6x-2y=2(3x-y)

b: =(x-y)(x-2)(x+2)

Lời giải:
a. Không phân tích được nữa

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

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

ko phân tích dc

b: =(x-y)(x-2)(x+2)

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

b) \(4x^2 -8x+3 \\= (2x^2)-2.2x .2 + 2^2 -1 \\=(2x-2)^2-1^2\\=(2x-2+1)(2x-2-1)\\= (2x-1)(2x-3)\)

ta có

\(5x=-3y=4z\)

\(\Rightarrow\frac{x}{12}=-\frac{y}{20}=\frac{z}{15}\)

\(\Rightarrow\frac{x}{12}=-\frac{y}{20}=\frac{3z}{45}=\frac{x-y+3z}{12+20+45}=\frac{7}{77}=\frac{1}{11}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{11}.12=\frac{12}{11}\\-y=\frac{1}{11}.20=\frac{20}{11}\\3z=\frac{1}{11}.45=\frac{45}{11}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{12}{11}\\y=-\frac{20}{11}\\z=\frac{45}{11}:3=\frac{15}{11}\end{cases}}\)

Vậy \(\hept{\begin{cases}x=\frac{12}{11}\\y=\frac{-20}{11}\\z=\frac{15}{11}\end{cases}}\)

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

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

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

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)

Cưu là mình vs (x^2+x)^2-2(x^2+x)-15