\(x^2-y^2+10x-6y+16\)

\(=\left(x^2+10x+25\right)-\left(y^2+6y+9\right)\)

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

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

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

d: \(=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)

\(10x\left(x-y\right)-6y\left(y-x\right)\)

\(=10x\left(x-y\right)+6y\left(x-y\right)\)

\(=\left(10x+6y\right)\left(x-y\right)\)

các bạn tl nhé minh cho

x^2-y^2+10x-6y-9

=x^2+10x -(y^2+6y+9)

= x^2+10x-(y^2+2.3.y+3^2)
=x^2+10x-(y+3)^2

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

={\(\left(x+x+3\right)\left[x-\left(x+3\right)\right]\)}\(+10x\)
=\(\left(x+x+3\right).\left(x-x-3\right)+10x\)

(đến đây b tự lm nhé, m cx k bt đúng k nx )

Ta có: \(\left(x+y\right)\left(y+z\right)\left(z+x\right)+xyz=x^2y+xy^2+xyz+y^2z+yz^2+xyz+xz^2+x^2x+xyz\)

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

\(x^2-y^2+10x-6y+16=\left(x^2+10x+25\right)-\left(y^2+6y+9\right)\)

\(=\left(x+5\right)^2-\left(y+3\right)^2=\left(x+y+8\right)\left(x-y+2\right)\)

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

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

phân tích đa thức thành nhân tử

a, 6x^2 + 7xy + 2y^2

=6x^2+3xy+4xy+2y^2

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

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

b, 9x^2 - 9xy - 4y^2

=9x^2 +3xy-12xy-4y^2

=3x(x+y)-4y(x+y)

=(3x+4y)(x+y)

c, x^2 - y^2 + 10x - 6y + 16=x^2-y^2+6x-6y+4x+16=x(x+6)-y(x+6)+4(x+6)=(x-y+4)(x+6)

Bài làm

a, 6x² + 7xy + 2y²

= 6x² + 3xy + 4xy + 2y² 

= ( 6x² + 3xy ) + ( 4xy + 2y² )

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

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

b, 9x² - 9xy - 4y² 

= 9x² - 12xy + 3xy - 4y² 

= ( 9x² - 12xy ) + ( 3xy - 4y² )

= 3x( 3x - 4y ) + y ( 3x - 4y )

= ( 3x + y )( 3x - 4y )

c, x² - y² + 10x - 6y + 16

 = x² - y² - 6x + 6y + 4x + 16

= x( x + 6 ) - y( x + 6 ) + 4( x + 6 )

= ( x - y + 4 )( x + 6 )

# Học tốt #

\(10x\left(x-y\right)-8y\left(y-x\right)\)

\(=10x\left(x-y\right)+8y\left(x-y\right)\)

\(=\left(x-y\right)\left(10x+8y\right)\)

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

 1/x-1/y=1/6 
<=> 6(y-x) = xy 
<=>y(6-x) = 6x (1) 
<=>y= 6x/(6-x) (2) 
(1) => x<6 => x=1,2,3,4,5 
Thay vào (2) ta có các cặp số nguyên thỏa đề bài là : 
(x;y)= (2;3);(3;6)

10x (x-y) -8y(y-x)

=10x(x-y) + 8y(x-y)

=(x-y)(10x+8y)

=2(x-y)(5x+4y)

a) 8x²-2x-1=8x²+4x-2x-2x-1=(8x²+4x)-(2x+1)=4x.(2x+1)-(2x+1)=(2x+1).(4x-1)

b) x²-y²+10x-6y+16x=x²-y²+10x+10y-16y+16x=(x-y)(x+y)+10(x+y)-16(y+x)=(x+y)(x-y+10+16)

câu b chắc là 16x chứ

`#3107.101107`

`x^2 - y^2 + 10x - 10y`

`= (x^2 - y^2) + (10x - 10y)`

`= (x - y)(x + y) + 10(x - y)`

`= (x + y + 10)(x - y)`

_____

Sử dụng HĐT:

`A^2 - B^2 = (A - B)(A + B).`