\(x^2-2xy+y^2+3x-3y-10=\left(x^2-2xy+y^2\right)+3\left(x-y\right)-10=\left(x-y\right)^2+3\left(x-y\right)-10=\left(x-y\right)^2-2\left(x-y\right)+5\left(x-y\right)-10=\left(x-y\right)\left(x-y-2\right)+5\left(x-y-2\right)=\left(x-y+5\right)\left(x-y-2\right)\)

= ( x - y)^2  - 3 ( x - y) . -10

= ( x  - y)^2 - 2.(x-y) . 3/2 +9/4 - 49/4 

= ( x - y - 3/2) ^2 - (7/2)^2 

= ( x- y - 3/2 - 7/2 )( x - y -3/2 + 7/2 )

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

x²-2xy+y²-3x-3y-10=(x-y)²

=(x-y)^2-3(x-y)-10

=(x-y)(x-y-3-10)

=(x-y)(x-y-13)

= ( x - y)^2  - 3 ( x - y) . -10

= ( x  - y)^2 - 2.(x-y) . 3/2 +9/4 - 49/4 

= ( x - y - 3/2) ^2 - (7/2)^2 

= ( x- y - 3/2 - 7/2 )( x - y -3/2 + 7/2 )

=( x - y - 5 )( x - y + 2) 

LÀm thế này đúng không cho nhận xét

trên là +3x mà sao dưới là -3x

\(x^2-2xy+y^2+3x-3y-10\)

\(=\left(y^2-xy-5y\right)-\left(xy-x^2-5x\right)+\left(2y-2x-10\right)\)

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

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

1) \(\left(x^2+8x+7\right).\left(x+3\right).\left(x+5\right)+15\)

\(=\left(x^2+8x+7\right).\left(x^2+5x+3x+15\right)+15\)

\(=\left(x^2+8x+7\right).\left(x^2+8x+15\right)+15\)

Ta đặt: \(x^2+8x+7=n\)

\(=n.\left(n+8\right)+15\)

\(=n^2+8n+15\)

\(=n^2+3n+5n+15\)

\(=\left(n^2+3n\right)+\left(5n+15\right)\)

\(=n.\left(n+3\right)+5.\left(n+3\right)\)

\(=\left(n+3\right).\left(n+5\right)\)

\(=\left(x^2+8x+7+3\right).\left(x^2+8x+7+5\right)\)

\(=\left(x^2+8x+10\right).\left(x^2+8x+12\right)\)

\(=\left(x^2+8x+10\right).\left(x^2+2x+6x+12\right)\)

\(=\left(x^2+8x+10\right).[x.\left(x+2\right)+6.\left(x+2\right)]\)

\(=\left(x^2+8x+10\right).\left(x+2\right).\left(x+6\right)\)

2) \(x^2-2xy+3x-3y-10+y^2\)

\(=\left(x-y\right)^2+3.\left(x-y\right)-10\)

Ta đặt: \(x-y=n\)

\(=n^2+3n-10\)

\(=n^2-2n+5n-10\)

\(=\left(n^2-2n\right)+\left(5n-10\right)\)

\(=n.\left(n-2\right)+5.\left(n-2\right)\)

\(=\left(n-2\right).\left(n+5\right)\)

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

e)  \(\left(x-y\right)^2+4\left(x-y\right)-12\)

\(=\left(x-y\right)\left[\left(x-y\right)+4\right]-12\)

\(=\left(x-y\right)\left(x-y+4\right)-12\)

f)  \(x^2-2xy+y^2+3x-3y-10\)

\(=\left(x^2-2xy+y^2\right)+\left(3x-3y\right)-10\)

\(=\left(x-y\right)^2+3\left(x-y\right)-10\)

\(=\left(x-y\right)\left[\left(x-y\right)+3\right]-10\)

\(=\left(x-y\right)\left(x-y+3\right)-10\)

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

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

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

\(=\left(x^2+4x+8\right)\left(x^2+7x+8\right)+2x^2\)

1) x(x+3)(x+1)(x+2)+1= (x^2 +3x)(x^2+3x+2)+1

Đặt x^2+3x=a ta có:

a(a+2)+1= a^2 +2a +1= (a+1)^2

Trở về ẩn x có

x(x+3)(x+1)(x+2)+1= (x^2 +3x)^2=x^2(x+3)^2

2) Đặt x^2 + x=a, ta có

a^2 +3a +2= (a^2+a) + (a+2)=a(a+2) +(a+2)=(a+1)(a+2)

Trở về ẩn x có

BT=( x^2 + x+1)(x^2 + x+2)

3) BT= (x-y)^2 +3(x-y) -10

đặt x-y=a ta có

a^2+3a -10= (a^2-2a)+(5a-10)=a(a-2)+5(a-2)=(a+5)(a-2)

trở về ẩn x,y có

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

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

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

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

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

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

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

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

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

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

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

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

\(=\left(x-y\right)\left[3-\left(x-y\right)\right]\)

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

a, ( x-y)²=4

3x^2 +3y^2 -6xy -12

=3(x^2 - 2xy +y^2 - 2^2  )

=3 (x-y)^2 - 2^2 

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

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

=3(x+y) -(x+y)^2 

(x+y)(3-x-y)