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

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

sao lại là x^2y

nghĩa là x^2 nhân với y

\(x^3-x^{22}-xy^2+y^2\)

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

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

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

\(x^3-x^2y-xy^2+y^3\)

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

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

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

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

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

Phạm Tuấn Đạt viết sai đề kìa

1.A

2.C

3.B

4.C

a

c

b

c

(x-1)y^2-4(x-1)y

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

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

3) \(xy\left(x+y\right)-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\) (xem lại đề sửa -2x thành -x mới đúng)

4) \(xy\left(x-3y\right)-2x+6y=xy\left(x-3y\right)-2\left(x-3y\right)=\left(x-3y\right)\left(xy-2\right)\)

Ta có: x² + y² - x²y² + xy - x - y

       = (x² - x²y²) + (y² - y) + (xy - x)

       = - x²(y² - 1) + y(y - 1) + x(y - 1)

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

       = (y - 1)(x + y - x²y - x²)

       = (y - 1)[- (x² - x) - (x²y - y)]

       = - (y - 1)[x(x - 1) + y(x² - 1)]

       = - (y - 1)[x(x - 1) + y(x + 1)(x - 1)]

       = - (y - 1)(x - 1)[x + y(x + 1)]

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

Ta có: x² + y² - x²y² + xy - x - y

       = (x² - x²y²) + (y² - y) + (xy - x)

       = - x²(y² - 1) + y(y - 1) + x(y - 1)

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

       = (y - 1)(x + y - x²y - x²)

       = (y - 1)[- (x² - x) - (x²y - y)]

       = - (y - 1)[x(x - 1) + y(x² - 1)]

       = - (y - 1)[x(x - 1) + y(x + 1)(x - 1)]

       = - (y - 1)(x - 1)[x + y(x + 1)]

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

Ai trên 10 điểm hỏi đáp thì mình nha mình đang cần gấp chỉ còn 59 điểm là tròn rồi mong các bạn hỗ trợ mình sẽ đền bù xứng đáng

Ta có: x² + y² - x²y² + xy - x - y

       = (x² - x²y²) + (y² - y) + (xy - x)

       = - x²(y² - 1) + y(y - 1) + x(y - 1)

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

       = (y - 1)(x + y - x²y - x²)

       = (y - 1)[- (x² - x) - (x²y - y)]

       = - (y - 1)[x(x - 1) + y(x² - 1)]

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

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

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

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

x(x-y)² -y(x-y)²+xy²-x²y=x(x-y)² -y(x-y)²+(xy²-x²y)=x(x-y)² -y(x-y)²+xy(x-y)=\(\left(x-y\right)\left[x\left(x-y\right)-y\left(x-y\right)+xy\right]\)=\(\left(x-y\right)\left[\left(x-y\right)^2+xy\right]\)