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

\(=\left[\left(4x^2\right)^2-1^2\right]\left(x-y\right)\)

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

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

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

Bài 1:

\(a,=3x\left(3xy+5y-1\right)\\ b,=\left(z-2\right)\left(3z-5\right)\\ c,=\left(x+2y\right)^2-4z^2=\left(x+2y+2z\right)\left(x+2y-2z\right)\\ d,=x^2-3x+5x-15=\left(x-3\right)\left(x+5\right)\)

Bài 2:

\(a,\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x+2-4x^2-12x=9\\ \Leftrightarrow4x^2+10x+7=0\\ \Leftrightarrow4\left(x^2+\dfrac{5}{2}x+\dfrac{25}{16}\right)+\dfrac{3}{4}=0\\ \Leftrightarrow4\left(x+\dfrac{5}{6}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\\ \Leftrightarrow x\in\varnothing\\ c,\Leftrightarrow x^2-12x+36=0\\ \Leftrightarrow\left(x-6\right)^2=0\\ \Leftrightarrow x=6\)

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

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

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

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

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

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

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

d) \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-8y^3=\left(x+1-2y\right)\left(x^2+2x+1+2xy+2y+4y^2\right)\)

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

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

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

c x^2(x+1)-4(x+1)

(x+1).(x^2-4)

d x^4-(x^2-2x+1)

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

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

MIK KO BIT DUNG HAY KO CON B THI MIK KO BIET LAM

Câu b dễ thôi

\(x^4-4x^3-8x^2+8x\)

\(=x\left(x^3-4x^2-8x+8\right)\)

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

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

b: \(=-xy^2\cdot x^2-xy^2\cdot z=-xy^2\left(x^2+z\right)\)

c: x^2y^3-xy^2

=xy^2*xy-xy^2

=xy^2(xy-1)

d: -x^3z-z

=z(-x^3-1)

=-z(x+1)(x^2-x+1)

e: =x(x-y)+(x-y)

=(x-y)(x+1)

n: =x^2(x-1)-(x-1)

=(x-1)(x^2-1)

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

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

2. x² - 2xy + y² - z² = (x - y)² - z² = (x - y - z)(x - y + z)

3. 5x - 5y + ax - ay = 5(x - y) + a(x - y) = (a + 5)(x - y)

4. a³ - a²x - ay + xy = a²(a - x) - y(a - x) = (a² - y)(a - x)

5. 4x² - y² + 4x + 1 = (2x + 1)² - y² = (2x + 1 - y)(2x  + y + 1)

6. x³ - x + y³ - y = (x + y)(x² - xy + y²) - (x + y) = (x + y)(x² - xy + y² - 1)

Trả lời:

1, x² - x - y² - y

= ( x² - y² ) - ( x + y )

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

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

2, x² - 2xy + y² - z²

= ( x² - 2xy + y² ) - z²

= ( x - y )² - x²

= ( x - y - z ) ( x - y + z )

3, 5x - 5y + ax - ay

= ( 5x + ax ) - ( 5y + ay )

= x ( 5 + a ) - y ( 5 + a )

= ( 5 + a ) ( x - y )

= ( 5 + a ) ( x - y )

4, a³ - a²x - ay + xy

= ( a³ - a²x ) - ( ay - xy )

= a² ( a - x ) - y ( a - x )

= ( a - x ) ( a² - y )

5, 4x² - y² + 4x + 1

= ( 4x² + 4x + 1 ) - y² 

= ( 2x + 1 )² - y²

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

6, x³ - x + y³ - y

= ( x³ + y³ ) - ( x + y )

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

= ( x + y ) ( x² - xy + y - 1 )