`= 36x^2-90x+225+15x-6x^2`

`= 30x^2-75x+225`

`= 15(2x^2-5x+9)`.

Đa thức này không phân tích được thành nhân tử.

bạn ghi lại cái đề cho rõ có đc ko?

$-25x^6-y^8+10x^3{y}^4$

$=-(25x^6-10x^3{y}^4+y^8)$

$=-[{\left(5x^3\right)}^2-2.5x^3{y}^4+{\left(y^4\right)}^2]$

$=-{(5x^3-y^4)}^2$

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

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

\(=\left(x^2+1-3\right)^2\)   

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

Trả lời:

( x² + 1 ) - 6 ( x² + 1 ) + 9

= ( x² + 1 ) - 2.( x² + 1 ).3 + 3²

=  [ ( x² + 1 ) - 3 ) ]²

= ( x² + 1 - 3 )²

= ( x² - 2 )²

a) \(x^2-xy+x-y\)

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

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

b) \(x^2+5x+6\)

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

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

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

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

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

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

Cưu là mình vs (x^2+x)^2-2(x^2+x)-15

Câu 1: C

Câu 2: A

Câu 3: B

Câu 4:B

Câu 5: A

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

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

\(=\left(x^2+5x\right)^2-36\)

Sửa đề: \(\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)-28\)

\(=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)-28\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)-28\)

\(=\left(x^2+5x\right)^2-64\)

\(=\left(x^2+5x+8\right)\left(x^2+5x-8\right)\)

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