a)2x^2+xy-y^2-x+2y-1

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

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

=2a^2+ab-b^2         với a=x,b=y-1

=2a^2+2ab-ab-b^2

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

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

a) Ta có: \(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)

\(\Leftrightarrow M=6x^2+9xy-y^2-5x^2+2xy\)

\(\Leftrightarrow M=x^2+11xy-y^2\)

Vậy: \(M=x^2+11xy-y^2\)

b) Ta có: \(\left(3xy-4y^2\right)-N=x^2-7xy+8y^2\)

\(\Leftrightarrow N=3xy-4y^2-x^2+7xy-8y^2\)

\(\Leftrightarrow N=-x^2+10xy-12y^2\)

Vậy: \(N=-x^2+10xy-12y^2\)

a, (6x²+9xy-y²) - ( 5x²-2xy)=M

=> M= (6x²+9xy-y²) - ( 5x²-2xy)

=> M= 6x²+9xy-y² - 5x²+2xy

=> M=(6x²- 5x²)+(9xy+2xy)-y²

=>M= 1x² + 11xy - y²

Vậy M= 1x² + 11xy - y²

b, N= (3xy-4y²) - (x²-7xy+8y²)

=> N= 3xy-4y² - x²+7xy-8y²

=> N= (3xy+7xy)-(4y²+8y²)-x²

=> N= 10xy - 12y² -x²

Vậy N= 10xy - 12y² -x²

a: Ta có: \(M+5x^2-2xy=6x^2+9xy-y^2\)

\(\Leftrightarrow M=6x^2+9xy-y^2-5x^2+2xy\)

\(\Leftrightarrow M=x^2+11xy-y^2\)

b: Ta có: \(\left(3xy-4y^2\right)-N=x^2-7xy+8y^2\)

\(\Leftrightarrow N=3xy-4y^2-x^2+7xy-8y^2\)

\(\Leftrightarrow N=-x^2+10xy-12y^2\)

x²-2xy+y²+3x-3y-10

= (x-y)²+3(x-y)-10

= [(x-y)²+5(x-y)]-[2(x-y)+10]

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

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

Ta có: \(x^2-2xy+y^2+3x-3y-10\)

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

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

a)\(5x^2-4\left(x^2-2x+1\right)-5=5\left(x^2-1\right)-4\left(x-1\right)^2=5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=\left(x-1\right)\left(5x+5-4x+4\right)=\left(x-1\right)\left(x+9\right)\)

b) \(9x^2+6x-4y^2+4y=\left(9x^2+6x+1\right)-\left(4y^2-4y+1\right)=\left(3x+1\right)^2-\left(2y-1\right)^2=\left(3x+1-2y+1\right)\left(3x+1+2y-1\right)=\left(3x-2y+2\right)\left(3x+2y\right)\)

a: \(5x^2-4\left(x^2-2x+1\right)-5\)

\(=5x^2-4x^2+8x-4-5\)

\(=x^2+8x-9\)

\(=\left(x+9\right)\left(x-1\right)\)

b: \(9x^2+6x-4y^2+4y\)

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

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

\(6x^2-13x+6\)

\(=6x^2-9x-4x+6\)

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

\(a.x^3-2x^2-2x-4\\ =\left(x^3-2x^2\right)-\left(2x-4\right)\\ =x^2\left(x-2\right)-2\left(x-2\right)\\ =\left(x^2-2\right)\left(x-2\right)\)

\(b.xy+1-x-y\\ =\left(xy-x\right)+\left(-y+1\right)\\ =x\left(y-1\right)-\left(y-1\right)\\ =\left(x-1\right)\left(y-1\right)\)

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

\(d.16-x^2+2xy-y^2\\ =4^2-\left(x-y\right)^2\\ =\left(4-x+y\right)\left(4-x-y\right)\)

b: =xy-x-y+1

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

=(x-1)(y-1)

c: =(x-2y)^2-4y

\(=\left(x-2y-2\sqrt{y}\right)\left(x-2y+2\sqrt{y}\right)\)

d: =16-(x^2-2xy+y^2)

=16-(x-y)^2

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

h: \(=\left(x+3\right)\cdot\left(x^2-3x+9\right)-4x\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2-7x+9\right)\)

Cái này y hệt cái đề mik thi:)

trời dài thế làm lâu phết đó nha hừm làm theo đúng công thức là được :)

b) 24x^2+6x^2y−2x−12yx−3y^2x+y 

tôi làm theo cách tìm tích số

nếu thấy đúng thì tick cho tôi nha

\(P-Q+R=2x^2-3xy+4y^2-3x^2-4xy+y^2+x^2+2xy+3y^2\)

\(P-Q+R=\left(2x^2-3x^2+x^2\right)-\left(3xy-4xy+2xy\right)+\left(4y^2+y^2+3y^2\right)\)

\(P-Q+R=x^2-1xy+8y^2\)