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

a) xy+3x-7y-21

=x(y+3)-7(x+3)

=(x-7)(y+3)

b)2xy-15-6x-5y

=2x(y-3)-5(-3+y)

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

c)2x^2y+2xy^2-2x-2y

=2x(xy-1)+2y(xy-1)

=(2x+2y)(xy-1)

x(x+3)-5x(x-5)-5(x+3)

=(x-5)(x+3)-5x(x-5)

=(x-5)(x+3-5x)

Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn

a) xy+3x-7y-21=(xy+3x)-(7y+21)= x(y+3)-7(y+3)=(y+3)(x-7)

b)2xy-15-6x+5y=(2xy-6x)+(5y-15)=2x(y-3)+5(y-3)=(y-3)(2x+5)

c)2x^2y+2xy^2-2x-2y=2xy(x+y)-2(x+y)=2(x+y)(xy-1)

d) x(x+3)-5x(x-5)-5(x+3)=[x(x+3)-5(x+3)]-5x(x-5)=(x+3)(x-5)-5x(x-5)=(x-5)(x+3-5x)=(x-5)(3-4x)

SORY I'M IN GRADE 6

Thợ Đào Mỏ Panda, mày bị điên à, không biết còn trả lời làm cái quái gì

Đã có đáp án:

2x^4-6x^3+x^2+6x-3

=2x^2-6x^3-3x^2-2x^2-6x-3

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

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

    \(2x^4+x^3-6x^2+x+2\) 

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

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

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

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

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

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

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

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

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

a) x³-2x²-x+2

=x(x²-1)+2(-x²+1)

=x(x²-1)-2(x²-1)

=(x²-1)(x-2)

b)

x²+6x-y²+9

=x²+6x+9-y²

=(x+3)²-y²

^{=(x+3-y)(x+3+y)}

1) \(x^3+2x-3\)

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

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

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

2) \(x^3-6x+4\)

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

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

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

3) \(x^3-2x^2+1\)

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

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

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

4) \(x^3+5x^2-12\)

\(=\left(x^3+2x^2\right)+\left(3x^2+6x\right)-\left(6x+12\right)\)

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

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

5) \(x^3-6x^2+9x\) (chắc đề như vậy)

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

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

6) \(4x^3-9x^2+5x\)

\(=x\left(4x^2-9x+5\right)\)

\(=x\left[\left(4x^2-4x\right)-\left(5x-5\right)\right]\)

\(=x\left[4x\left(x-1\right)-5\left(x-1\right)\right]\)

\(=x\left(x-1\right)\left(4x-5\right)\)

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

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

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

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

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

\(x^4+2x^3+6x-9=0\)\(\)

\(\Rightarrow x^4+3x^3-x^3+9x-3x-9+3x^2-3x^2=0\)

\(\Rightarrow\left(x^4+3x^3+3x^2+9x\right)-\left(x^3+3x^2+3x+9\right)=0\)

\(\Rightarrow x\left(x^3+3x^2+3x+9\right)-\left(x^3+3x^2+3x+9\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x^3+3x^2+3x+9\right)=0\)

\(\Rightarrow\left(x-1\right)\left[x^2\left(x+3\right)+3\left(x+3\right)\right]=0\)

\(\Rightarrow\left(x-1\right)\left(x^2+3\right)\left(x+3\right)=0\)

Mà \(x^2+3>0\) với mọi x\(\in\)R.

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)