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)

\(3,\)Nhẩm nghiệm của đa thức trên ta đc : -1

Ta có lược đồ sau :

Phân tích thành nhân tử ta có :\(\left(x+1\right)\left(x^2-4\right)\)

Lời giải:
$\frac{x}{y}$ không phải đơn thức bạn nhé.

a. $x^2-2x+1=(x-1)^2$

b. $x^2+2xy-25+y^2=(x^2+2xy+y^2)-25=(x+y)^2-5^2=(x+y-5)(x+y+5)$

c. $5x^2-10xy=5x(x-2y)$

d. $x^2-y^2+x-y=(x^2-y^2)+(x-y)=(x-y)(x+y)+(x-y)$

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

a: \(A=x\sqrt{x}-y\sqrt{y}+x\sqrt{y}-y\sqrt{x}\)

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

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

b: \(B=5x^2-7x\sqrt{y}+2y\)

\(=5x^2-5x\sqrt{y}-2x\sqrt{y}+2y\)

\(=5x\left(x-\sqrt{y}\right)-2\sqrt{y}\left(x-\sqrt{y}\right)\)

\(=\left(x-\sqrt{y}\right)\left(5x-2\sqrt{y}\right)\)

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

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

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

8) \(6x^2y+4xy^2+2xy=2xy\left(3x+2y+1\right)\)

9) \(5x^2\left(x-2y\right)-15x\left(x-2y\right)=5x\left(x-2y\right)\left(x-3\right)\)

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

6) 9x³y² + 3x²y² = 3x²y²( 3x + 1 )

7) x³ + 2x² + 3x = x( x² + 2x + 3 )

8) 6x²y + 4xy² + 2xy = 2xy( 3x + 2y + 1 )

9) 5x²( x - 2y ) - 15x( x - 2y ) = 5x( x - 2y )( x - 3 )

10 3( x - y ) - 5x( y - x ) = 3( x - y ) + 5x( x - y ) = ( x - y )( 3 + 5x )

a, \(9x^3y^2+3x^2y^2=3x^2y^2\left(3x+1\right)\)

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

c, \(6x^2y+4xy^2+2xy=2xy\left(3x+2y+1\right)\)

d, \(5x^2\left(x-2y\right)-15x\left(x-2y\right)=\left(5x^2-15x\right)\left(x-2y\right)=5x\left(x-3\right)\left(x-2y\right)\)

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

a,(x+y)(2a-4)

b,(x+y)(a-b)

c,a(b+a)(x-5)

d,2a(a+2)(x+y)

**** cho mk nha

câu a ấy, làm chưa xong à?

a) 2a ( x + Y ) - 4 ( x + y )

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

b) x ( a - b ) - y ( b - a )

= x ( a - b ) +  y ( a - b)

= ( a - b ) ( x + y )

c) ab ( x - 5 ) - a^2 ( 5 - x )

= ab ( x - 5 ) + a^2 ( x - 5 )

= ( x - 5 ) ( ab + a^2 )

= ( x - 5 ) a ( b + a )

d) 2a^2 ( x - y) - 4a ( y - x )

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

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

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

đầu cắt moi :)))))

11) \(3x\left(x-1\right)+5\left(1-x\right)=\left(3x-5\right)\left(x-1\right)\)

12) \(2\left(2x-1\right)+3\left(1-2x\right)=1-2x\)

13) \(10x\left(x-y\right)-8y\left(y-x\right)=2\left(x-y\right)\left(5x+4y\right)\)

14) \(3x\left(y+2\right)-3\left(y+2\right)=3\left(x-1\right)\left(y+2\right)\)

15) \(x^2-y^2-2x+2y\)

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

m có bị điên ko mà trọc bn ấy?!!!!

\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)

\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11

e: Ta có: \(x^2-6x+y^2+4y+2=0\)

\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)

Dấu '=' xảy ra khi x=3 và y=-2