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

Đặt t = x²  + 3x + 1

Khi đó PT có dạng:

t.(t + 1) - 6

= t² + t - 6

= t² - 2t - 3t - 6

= t.(t - 2) + 3.(t - 2)

= (t + 3).(t - 2)

= (x² + 3x + 1 + 3).(x² + 3x + 1 - 2)

= (x² + 3x + 4).(x² + 3x - 1)

\(1\hept{\begin{cases}\left(x^2+3x+2-1\right)\left(x^2+2x+2\right)-6\\\left(t-1\right)\left(t\right)-6\\t^2-t-6\end{cases}}.\) " đặt x^2+3x+2 = t

\(\hept{\begin{cases}t^2-\frac{2t.1}{2}+\frac{1}{4}-\left(\frac{24+1}{4}\right)\\\left(t-\frac{1}{2}\right)^2-\frac{25}{4}\\\left(t-\frac{1}{2}\right)^2-\frac{25}{4}\end{cases}}\)

\(\hept{\begin{cases}\left(t-\frac{1}{2}-\frac{5}{2}\right)\left(t-\frac{1}{2}+\frac{5}{2}\right)\\\left(t-\frac{7}{2}\right)\left(t+\frac{4}{2}\right)\\\left(t-\frac{7}{2}\right)\left(t+\frac{4}{2}\right)\end{cases}}\)

2)  \(\hept{\begin{cases}\left\{\left(x+1\right)\left(x+7\right)\right\}\left\{\left(x+5\right)\left(x+3\right)\right\}+15\\\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\\t\left(t+8\right)+15\end{cases}}\)  

\(\hept{\begin{cases}t^2+8t+15\\\left(t^2+8t+16\right)-1\\\left(t+4\right)^2-1\end{cases}}\Leftrightarrow\left(t+5\right)\left(t+4\right)\)

\(\hept{\begin{cases}a^3\left(b-c\right)+b^3\left(c-a+b-b\right)+c^3\left(a-b\right)\\a^3\left(b-c\right)-b^3\left(-c+a-b+b\right)+c^3\left(a-b\right)\\a^3\left(b-c\right)-b^3\left(a-b\right)-b^3\left(b-c\right)+c^3\left(a-b\right)\end{cases}\Leftrightarrow\hept{\begin{cases}\left(b-c\right)\left(a^3-b^3\right)-\left(a-b\right)\left(b^3-c^3\right)\\\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)-\left(a-b\right)\left(b-c\right)\left(b^2+ab+c^2\right)\\\left(a-b\right)\left(b-c\right)\left(a^2+2ab+2b^2+c^2\right)\end{cases}}}\)

a) \(x^2-xy+x-y\)

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

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

b) \(x^2+5x+6\)

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

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

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

\(2xy-x^2-y^2+16\)

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

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

1. \(\left(x+1\right)^3-125\)

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

\(=\left(x+1-5\right).\left[\left(x+1\right)^2+\left(x+1\right).5+5^2\right]\)

2. \(\left(x+4\right)^3-64\)

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

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

3. \(x^3-\left(y-1\right)^3\)

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

\(\)4. \(\left(a+b\right)^3-c^3\)

\(=\left[\left(a+b\right)-c\right].\left[\left(a+b\right)^2+\left(a+b\right).c+c^2\right]\)

5. \(125-\left(x+2\right)^3\)

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

\(=\left(5-x-2\right).\left[5^2+5.\left(x+2\right)+\left(x+2\right)^2\right]\)

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

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

1

a, 2x²+4x+2-2y² = 2(x²+2x+1-y²)= 2[(x+1)²-y² ] = 2(x-y+1)(x+y+1)

b, 2x - 2y - x² + 2xy - y²= 2(x -y) - (x² - 2xy + y²) = 2(x-y)-(x-y)²=(x-y)(2-x+y)

c, x²-y²-2y-1=x²-(y²+2y+1)=x²-(y+1)²=(x-y-1)(x+y+1)

d, x²-4x-2xy-4y+y²= x²-2xy+y²-4x-4y=(x-y)

2.

a, x²-3x+2=x²-x-2x+2=x(x-1)-2(x-1)=(x-2)(x-1)

b, x²+5x+6=x²+2x+3x+6=x(x+2)+3(x+2)=(x+3)(x+2)

c, x²+6x-6=

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

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

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

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

b) \(x^3-x+3x^2y+3xy^2+y^3-y\)

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

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

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

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

Bài 1 yêu cầu gì em?

Bài 2:

\(a,x\left(x-1\right)+5\left(x-1\right)=\left(x+5\right)\left(x-1\right)\\ b,3x\left(x+1\right)+3\left(x+1\right)=\left(3x+3\right)\left(x+1\right)=3\left(x+1\right)\left(x+1\right)=3\left(x+1\right)^2\\ c,x\left(x-3\right)+xy\left(x-3\right)=\left(x+xy\right)\left(x-3\right)=x\left(y+1\right)\left(x-3\right)\\ d,2x\left(x-2\right)-6\left(x-2\right)=\left(2x-6\right)\left(x-2\right)=2\left(x-3\right)\left(x-2\right)\)

Bài 1:

a) \(3xy+6y\)

\(=3y\left(x+2\right)\)

b) \(3x^2+9x\)

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

c) \(6x-9y^2\)

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

d) \(10xy^2-6x^2y\)

\(=2xy\left(5y-3x\right)\)

Bài 2:

a) \(x\left(x-1\right)+5\left(x-1\right)\)

\(=\left(x-1\right)\left(x+5\right)\)

b) \(3x\left(x+1\right)+3\left(x+1\right)\)

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

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

\(=3\left(x+1\right)^2\) 

c) \(x\left(x-3\right)+xy\left(x-3\right)\)

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

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

d) \(2x\left(x-2\right)-6\left(x-2\right)\)

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

\(=2\left(x-3\right)\left(x-2\right)\)

Bài `1`

`a,3xy +6y`

`= 3y(x+2)`

`b,3x^2+9x`

`= 3x(x+3)`

`c,6x-9y^2`

`= 3(2x- 3y^2)`

`d,10xy^2-6x^2y`

`= 2xy(5y-3x)`

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

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

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

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)