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a) \(=2\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
\(=2\left(x-y\right)-\left(x-y\right)^2\)
\(=\left(x-y\right)\left(2-x+y\right)\)
b) \(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+y^3\right)+\left(3x^2+3xy^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2+3xy-1\right)\)
\(=\left(x+y\right)\left(x^2+y^2+2xy-1\right)\)
1
a, 2x2+4x+2-2y2 = 2(x2+2x+1-y2)= 2[(x+1)2-y2 ] = 2(x-y+1)(x+y+1)
b, 2x - 2y - x2 + 2xy - y2= 2(x -y) - (x2 - 2xy + y2) = 2(x-y)-(x-y)2=(x-y)(2-x+y)
c, x2-y2-2y-1=x2-(y2+2y+1)=x2-(y+1)2=(x-y-1)(x+y+1)
d, x2-4x-2xy-4y+y2= x2-2xy+y2-4x-4y=(x-y)
2.
a, x2-3x+2=x2-x-2x+2=x(x-1)-2(x-1)=(x-2)(x-1)
b, x2+5x+6=x2+2x+3x+6=x(x+2)+3(x+2)=(x+3)(x+2)
c, x2+6x-6=
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
a: \(=x^2\left(x-y\right)+2014\left(x-y\right)=\left(x-y\right)\left(x^2+2014\right)\)
a) xy+3x-7y-21
=x(y+3)-7(x+3)
=(x-7)(y+3)
b)2xy-15-6x-5y
=2x(y-3)-5(-3+y)
=(2x-5)(y-3)
c)2x^2y+2xy^2-2x-2y
=2x(xy-1)+2y(xy-1)
=(2x+2y)(xy-1)
x(x+3)-5x(x-5)-5(x+3)
=(x-5)(x+3)-5x(x-5)
=(x-5)(x+3-5x)
Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a: \(x\left(2x-y\right)-y\left(2x-y\right)=\left(2x-y\right)\left(x-y\right)\)
c: \(x^2-3x+3y-y^2\)
\(=\left(x-y\right)\left(x+y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-3\right)\)
b: \(x^2-6x-7=\left(x-7\right)\left(x+1\right)\)
a) \(x\left(2x-y\right)-y\left(2x-y\right)=\left(2x-y\right)\left(x-y\right)\)
b) \(x^2-6x-7=x\left(x-7\right)+\left(x-7\right)=\left(x-7\right)\left(x+1\right)\)
c) \(x^2-3x+3y-y^2=\left(x-y\right)\left(x+y\right)-3\left(x-y\right)=\left(x-y\right)\left(x+y-3\right)\)
d) \(x^3-xy+2y-8=\left(x-2\right)\left(x^2+2x+4\right)-y\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+4-y\right)\)
\(x^2-y^2-2x+2y=\left(x+y\right)\left(x-y\right)-2\left(x-y\right)=\left(x+y-2\right)\left(x-y\right)\) \(2x+2y-x^2-xy=2\left(x+y\right)-x\left(x+y\right)=\left(2-x\right)\left(x+y\right)\)
\(3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2-4c^2\right)=3\left(a-b\right)^2-3\left(2c\right)^2=3\left(a-b+2c\right)\left(a-b-2c\right)\) \(x^2-25+y^2+2xy=\left(x+y\right)^2-25=\left(x+y-5\right)\left(x+y+5\right)\)
\(a^2+2ab+b^2-ac-bc=\left(a+b\right)^2-c\left(a+b\right)=\left(a+b\right)\left(a+b-c\right)\)
\(x^2-2x+1-4y^2-4y-1=\left(x-1\right)^2-\left(2y+1\right)^2=\left(x+2y\right)\left(x-2y-2\right)\)\(x^2\left(x-1\right)-16\left(x-1\right)=\left(x-1\right)\left(x^2-16\right)=\left(x-1\right)\left(x+4\right)\left(x-4\right)\) \(x^2y-x^3-9y+9x=x^2\left(y-x\right)-9\left(y-x\right)=\left(y-x\right)\left(x^2-9\right)\left(y-x\right)\left(x-3\right)\left(x+3\right)\)