\(1,\\ a,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ b,=a^2\left(a-x\right)-y\left(a-x\right)=\left(a^2-y\right)\left(a-x\right)\\ c,=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\\ d,=x\left(x-2y\right)+t\left(x-2y\right)=\left(x+t\right)\left(x-2y\right)\\ 2,\\ \Rightarrow x^2-4x+4-x^2+9=6\\ \Rightarrow-4x=-7\Rightarrow x=\dfrac{7}{4}\\ 3,\\ a,x^2+2x+2=\left(x+1\right)^2+1\ge1>0\\ b,-x^2+4x-5=-\left(x-2\right)^2-1\le-1< 0\)

\(a,36-4a^2+20ab-25b^2\)

\(=6^2-\left(2a-5b\right)^2=\left(6-2a+5b\right)\left(6+2a-5b\right)\)\(b,x^2+2xy+y^2-xz-yz\)

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

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

\(d,5a^2-10a^2b+5ab^2-10a+10b\)

\(=5a^2-5a^2b-5a^2b+5ab^2-10a+10b\)

\(=5a\left(a-b\right)-5ab\left(a-b\right)-10\left(a-b\right)\)

\(=\left(a-b\right)\left(5a-5ab-10\right)\)

c) x2 + 2xy + y2 – xz – yz = (x + y)2 – z(x + y) = (x + y)(x + y – z)

c) x2 + y2 + xz + yz + 2xy

= (x2 + 2xy + y2) + (xz + yz)

= (x + y)2 + z(x + y)

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

Bạn thử xem lại đề câu d nhé.

a) Ta có: \(4x\left(2x-3y\right)-8y\left(3y-2x\right)\)

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

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

b) Ta có: \(4x^2-4xy+y^2-9z^2\)

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

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

c) Ta có: \(x^2y+yz+xy^2+xz\)

\(=xy\left(x+y\right)+z\left(x+y\right)\)

\(=\left(x+y\right)\left(xy+z\right)\)

d) (8a3 – 27b3) – 2a(4a2 – 9b2)

= (2a – 3b)(4a2 + 6ab + 9b2) – 2a(2a – 3b)(2a + 3b)

= (2a – 3b)(4a2 + 6ab + 9b2 – 4a2 – 6ab) = 9b2(2a – 3b)

a)\(36-4a^2+20ab-25b^2=6^2-\left(4a^2-20ab+25b^2\right)\)

\(=6^2-\left[\left(2a\right)^2-2.2a.5b+\left(5b\right)^2\right]\)

\(=6^2-\left(2a-5b\right)^2\)

\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)

b)\(a^3+3a^2+3a+1-27b^3=\left(a+1\right)^3-\left(3b\right)^3\)(chỗ này mình sửa 27b² thành 27b³ vì mình nghĩ nhầm đề)

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

\(=\left(a+1-3b\right)\left(a^2+2a+1+3ab+3b+9b^2\right)\)

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

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

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

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

a)  36-4a²+20ab-25b²

= 6^2 - (4a^2 - 20xb + 25b^2)

= 6^2 - (2a - 5b)^2

= [6 - (2a - 5b)] [6 + (2a - 5b)]

= (6 - 2a + 5b) (6 + 2a -5b)

a) \(6^2-\left(\left(2a\right)^2-2.2a.5b+\left(5b\right)^2\right)\)

\(=6^2-\left(2a-5b\right)^2\)

\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)

b) \(=\left(a+1\right)^3-\left(3b\right)^3\)

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

\(=\left(a+1-3b\right)\left(a^2+2a+1+3ab+3b+9b^2\right)\)

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

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

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

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

a) \(3xy^2-12xy+12x\)

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

b) \(3x^3y-6x^2y-3xy^3-6axy^2-3a^2xy+3xy\)

\(=3xy\left(x^2-2x-y^2-2ay-a^2+1\right)\)

\(=3xy\left[\left(x^2-2\cdot x\cdot1+1^2\right)-\left(y^2+2\cdot y\cdot a+a^2\right)\right]\)

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

\(=3xy\left(x-1-y-a\right)\left(x-1+y+a\right)\)

c) \(36-4a^2+20ab-25b^2\)

\(=6^2-\left[\left(2a\right)^2-2\cdot2a\cdot5b+\left(5b\right)^2\right]\)

\(=6^2-\left(2a-5b\right)^2\)

\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)

d) \(5a^3-10a^2b+5ab^2-10a+10b\)

\(=5a\left(a^2-2ab+b^2\right)-10\left(a-b\right)\)

\(=5a\left(a-b\right)^2-10\left(a-b\right)\)

\(=\left(a-b\right)\left[5a\left(a-b\right)-10\right]\)

\(=5\left(a-b\right)\left[a\left(a-b\right)-2\right]\)

\(=5\left(a-b\right)\left(a^2-ab-2\right)\)

a. 3xy²-12xy+12x

= 3x(y²-4y+4)

= 3x(y-2)²

b. 3x³y-6x²y-3xy³-6axy²-3a²xy+3xy

= 3xy( x²-2x-y²-2ay-a²+1)

= 3xy ((x²-2x+1)-(a²-2ay-y²))

=3xy((x-1)²-(a-y)²)

= 3xy((x-1+a-y)(x-1-(a-y))

=3xy(x-1+a-y)(x-1-a+y)

d. =( 5a(a²-2ab+b²))-(10(a+b))

= 5a(a-b)²-10(a-b)

=5a(a-b)(a-b)-10(a-b)

=(a-b)(5a(a-b)-10)

Hình như mik làm sai hết rồi

a) Cách 1.

Ta có 2xy + 3z + 6y + xz = (2xy + xz) + (3z + 6y)

= x(2 y + z)+3(z + 2 y) = (z + 2y)(x + 3).

Cách 2.

Ta có 2xy + 3z + 6y + xz = (2x1/ + 6y) + (3z + xz)

= 2y(x + 3) + z(3 + x) = (z + 2y)(x + 3).

b) Biến đổi được a 4   -   9 rt 3   +   a 2 -9a = (a- 9)a( a 2  +1).

c) Biến đổi được 3 x 2  + 5y - 3xy + (-5x) = (x - y)(3x - 5).

d) Biến đổi được  x 2  - (a + b)x + ab = (x- a)(x - b).

e) Ta có 4 x 2 - 4xy + y 2   –   9 t 2 =  ( 2 x   -   y ) 2   -   ( 3 t ) 2

= (2x - y - 3t )(2x - y + 31).

g) Ta có  x 3   -   3 x 2 y   +   3 xy 2   -   y 3   -   z 3

= ( x   -   y ) 3   -   z 3 = (x - y - z)( x 2   +   y 2   +   z 2  - 2xy + xz - yz).

h) Ta có x 2   -   y 2 + 8x + 6y+ 7 = ( x 2  +8x + 16) - ( y 2  - 6y+ 9)

= ( x   +   4 ) 2   - ( y - 3 ) 2  =(x-y + 7)(x + y + l).