\(=4x^2y^2+2xy-6-4x^2y^2+4xy-1=6xy-7=6\cdot\dfrac{1}{3}\cdot3-7=6-7=-1\)

a) \(3x^2-3xy-5x+5y\)

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

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

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

b) \(2x^3y-2xy^3-4xy^2-2xy\)

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

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

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

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

c) \(x^2+1+2x-y^2\)

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

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

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

d) \(x^2+4x-2xy-4y+y^2\)

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

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

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

e) \(x^3-2x^2+x\)

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

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

f) \(2x^2+4x+2-2y^2\)

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

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

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

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

a: =3x(x-y)-5(x-y)

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

b: \(=2xy\left(x^2-y^2-2y-1\right)\)

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

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

d:

Sửa đề: x^2+4x-2xy-4y+y^2

=x^2-2xy+y^2+4x-4y

=(x-y)^2+4(x-y)

=(x-y)(x-y+4)

e: =x(x^2-2x+1)

=x(x-1)^2

f: =2(x^2+2x+1-y^2)

=2[(x+1)^2-y^2]

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

a: =(xy-2x)-(y^2-2y)

=x(y-2)-y(y-2)

=(x-y)(y-2)

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

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

=(x-y)(x-y-1)

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

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

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

d: =(x+3)(x+3-2x+5)

=(x+3)(8-x)

\(a,xy-2x-y^2+2y\)

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

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

\(b,x^2-2xy+y^2-x+y\)

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

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

\(c,x^2-1-2xy+2y\)

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

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

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

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

\(=\left(x+3\right)\left(-x+8\right)\)

#Urushi

a) ( x - 1 )³ + 3x( x - 1 )² + 3x²( x - 1 ) + x³

= [ ( x - 1 ) + x ) ]³ ( HĐT số 4 )

= [ x - 1 + x ]³

= [ 2x - 1 ]³ 

=> đpcm

b) ( x² - 2xy )³ + 3( x² - 2xy )y² + 3( x² - 2xy )y⁴ + y⁶

= [ ( x² - 2xy ) + y² ]³ ( HĐT số 4 )

= [ x² - 2xy + y² ]³

= [ ( x - y )² ]³

= ( x - y )⁶

=> đpcm

`Answer:`

\(a)\left(-3x^2y-2xy^2+6\right)+\left(-x^2y+5xy^2-1\right)\)

\(=-3x^2y-2xy^2+6+-x^2y+5xy^2-1\)

\(=\left(-3x^2y-x^2y\right)+\left(-2xy^2+5xy^2\right)+\left(6-1\right)\)

\(=-4x^2y+3xy^2+5\)

\(b)\left(1,6x^3-3,8x^2y\right)+\left(-2,2x^2y-1,6x^3+0,5xy^2\right)\)

\(=1,6x^3-3,8x^2y+-2,2x^2y-1,6x^3+0,5xy^2\)

\(=\left(1,6x^3-1,6x^3\right)+\left(-3,8x^2y+-2,2x^2y\right)+0,5xy^2\)

\(=-6x^2y+0,5xy^2\)

\(c)\left(6,7xy^2-2,7xy+5y^2\right)-\left(1,3xy-3,3xy^2+5y^2\right)\)

\(=6,7xy^2-2,7xy+5y^2-1,3xy+3,3xy^2-5y^2\)

\(=\left(6,7xy^2+3,3xy^2\right)+\left(-2,7xy-1,3xy\right)+\left(5y^2-5y^2\right)\)

\(=10xy^2+-4xy\)

\(=10xy^2-4xy\)

\(d)\left(3x^2-2xy+y^2\right)+\left(x^2-xy+2y^2\right)-\left(4x^2-y^2\right)\)

\(=3x^2-2xy+y^2+x^2-xy+2y^2-4x^2+y^2\)

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

\(=-3xy+4y^2\)

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

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

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

\(=-1\)

Câu a pt đầu là \(x^2+2xy^2=3\) hay \(x^3+2xy^2=3\) vậy nhỉ? Nhìn \(x^2\) chẳng hợp lý chút nào

b. \(\Leftrightarrow\left\{{}\begin{matrix}x^2\left(xy+1\right)-y\left(xy+1\right)+xy+1=2\\\left(x^4+y^2-2x^2y\right)+xy+1=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-y\right)\left(xy+1\right)+xy+1=2\\\left(x^2-y\right)^2+xy+1=2\end{matrix}\right.\)

Trừ vế cho vế:

\(\left(x^2-y\right)\left(xy+1\right)-\left(x^2-y\right)^2=0\)

\(\Leftrightarrow\left(x^2-y\right)\left(xy+1-x^2+y\right)=0\)

\(\Leftrightarrow\left(x^2-y\right)\left[y\left(x+1\right)+\left(x+1\right)\left(1-x\right)\right]=0\)

\(\Leftrightarrow\left(x^2-y\right)\left(x+1\right)\left(y+1-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=x^2\\x=-1\\y=x-1\end{matrix}\right.\)

- Với \(y=x^2\) thế xuống pt dưới:

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

....

Hai trường hợp còn lại bạn tự thế tương tự

\(2x^3y-2xy^3-4xy^2-2xy\)

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

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

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

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

Vậy chọn đáp án A

chọn A

a) ( x + 2 )( x² - 2x + 4 ) - ( 18 + x³ )

= x³ + 8 - 18 - x³ = -10

b) ( 2x - y )( 4x² + 2xy + y² ) - ( 2x + y )( 4x² - 2xy + y² )

= 8x³ - y³ - ( 8x³ + y³ )

= 8x³ - y³ - 8x³ - y³ = -2y³

c) ( x - 3 )( x + 3 ) - ( x + 5 )( x - 1 )

= x² - 9 - ( x² + 4x - 5 )

= x² - 9 - x² - 4x + 5 = -4x - 4

d) ( 3x - 2 )² + ( x + 1 )² + 2( 3x - 2 )( x + 1 )

= ( 3x - 2 + x + 1 )² 

= ( 4x - 1 )²