Bài 4.

a) 3xy² - 45x²y = 3xy( y - 15x )

b) 25y² - 4x² + 4x - 1

= 25y² - ( 4x² - 4x + 1 )

= ( 5y )² - ( 2x - 1 )²

= ( 5y - 2x + 1 )( 5y + 2x - 1 )

c) x² - 5x + xy - 5y

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

= ( x - 5 )( x + y )

d) x² - 8x - 33

= x² + 3x - 11x - 33

= x( x + 3 ) - 11( x + 3 )

= ( x + 3 )( x - 11 )

Bài 5.

a) A = ( x - 2 )³ - x²( x - 4 ) + 8

= x³ - 6x² + 12x - 8 - x³ + 4x² + 8

= -2x² + 12x

B = ( x² - 6x + 9 ) : ( x - 3 ) - x( x + 7 ) - 9

= ( x - 3 )² : ( x - 3 ) - x² - 7x - 9

= x - 3 - x² - 7x - 9

= -x² - 6x - 12

b) Với x = -1 thì A = -2.(-1)² + 12.(-1) = -2 - 12 = -14

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

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

b: =(2x-5y)(2x+5y)

c: =x(x^3-8)

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

a)

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

b)

\(4x^2-25y^2\\ =\left(2x-5y\right)\left(2x+5y\right)\)

c)

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

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

\(b,4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)

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

Phân tích các đa thức sau thành nhân tử ... c) 6x(x+y)^2+3x^2y(x+y). 2: .... x³ - 5x + 8x - 4=x² . x -5x + 8x -2² = (x² - 2²) . (x -5x + 8x )=(x-2) . (x+2) . 4x. x³ - 9x² ..... Phân tích các đa thức sau thành nhân tử : a,x^3+5x^2+8x+4 b, x^3-9x^2+6x+16 .

\(a.=x^2\left(x-y\right)-25\left(x-y\right)\)

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

\(b.=\left(b-a\right)\left(b+a\right)-4\left(b-a\right)\)

\(=\left(b+a-4\right)\left(b-a\right)\)

a ) x³ - x²y - 25x + 25y

= x² . ( x - y ) - 25 . ( x - y )

= ( x - y )( x² - 25 )

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

b ) b² - 4b + 4a - a²

= ( b² - a² ) - ( 4b - 4a )

= ( b - a )( a + b ) - 4 . ( b - a )

= ( b - a )( a + b - 4 )

c ) x³ - 4x² - 8x + 8

= ( x³ + 8 ) - ( 4x² + 8x )

= ( x + 2 )( x² - 2x + 4 ) - 4x . ( x + 2 )

= ( x + 2 )( x² - 2x + 4 - 4x )

= ( x + 2 )( x² - 6x + 4 )

\(1,a^2-b^2-12a+12b=\left(a-b\right)\left(a+b\right)-12\left(a-b\right)=\left(a-b\right)\left(a+b-12\right)\\ 2,4x^2-4x+1-25y^2=\left(2x-1\right)^2-\left(5y\right)^2=\left(2x-5y-1\right)\left(2x+5y-1\right)\\ c,x^2-3x-10=\left(x^2-5x\right)+\left(2x-10\right)=x\left(x-5\right)+2\left(x-5\right)=\left(x-5\right)\left(x+2\right)\)

a,\(x^2+4x-25y^2+4=\left(x^2+4x+4\right)-\left(5y\right)^2=\left(x+2\right)^2-\left(5y\right)^2\)

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

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

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

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

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

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

Mk làm sai câu b nha , phảo làm như sau 

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

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

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

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

\(a,=2xy\left(2y-x\right)\\ b,=x^2\left(x-4\right)+5\left(x-4\right)=\left(x^2+5\right)\left(x-4\right)\\ c,=\left(x-y\right)\left(x^2-25\right)=\left(x-y\right)\left(x-5\right)\left(x+5\right)\)

a: 6x-2y=2(3x-y)

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

Lời giải:
a. Không phân tích được nữa

b. $x^2(x-y)+4(y-x)=x^2(x-y)-4(x-y)=(x-y)(x^2-4)=(x-y)(x-2)(x+2)$
c. $x^3+2x^2y+xy^2-4x=x(x^2+2xy+y^2-4)$

$=x[(x^2+2xy+y^2)-4]=x[(x+y)^2-2^2]=x(x+y-2)(x+y+2)$

ko phân tích dc

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

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

a) Ta có: \(x^2-2xy+y^2-2x+2y\)

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

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

b) Ta có: \(x^2-4x+4-x^2y+2xy\)

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

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

c) Ta có: \(ax^2-3axy-x^2+6xy-9y^2\)

\(=ax\left(x-3y\right)-\left(x^2-6xy+9y^2\right)\)

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

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

d) Ta có: \(2a^2x-5a^2y-4x^2+30xy-25y^2\)

\(=a^2\left(2x-5y\right)-\left(4x^2-30xy+25y^2\right)\)

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

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