a) \(x^2-xy+x-y\)

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

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

b) \(x^2+5x+6\)

\(=x^2+2x+3x+6\)

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

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

\(2xy-x^2-y^2+16\)

\(=16-\left(x-y\right)^2\)

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

1: =(x+y-3x)(x+y+3x)

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

2: =(3x-1-4)(3x-1+4)

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

=3(x+1)(3x-5)

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

=-[(x^2+1)^2-(2x)^2]

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

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

4: =(2x+1+x-1)(2x+1-x+1)

=3x(x+2)

5: =[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]

=(2x^2+2)*4x

=8x(x^2+1)

6: =(5x-5y)^2-(4x+4y)^2

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

=(x-9y)(9x-y)

7: =(x^2+xy+y^2+xy)(x^2+xy-y^2-xy)

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

=(x+y)^3*(x-y)

8: =(x^2+4y^2-20-4xy+16)(x^2+4y^2-20+4xy-16)

=[(x-2y)^2-4][(x+2y)^2-36]

=(x-2y-2)(x-2y+2)(x+2y-6)(x+2y+6)

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

Bài làm

   -x² - y² + xy + 16

= -( x² - xy + y² ) + 16

= -( x - y )² + 4² 

= -[ ( x - y )² - 4² ]

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

# Học tốt #

a) x2 + xy –x – y = x(x + y) – (x + y) = (x + y)(x -1 ).

b) a2 – b2 + 8a + 16 = (a2 + 8a + 16) – b2 = (a + 4)2 – b2

= (a + 4 – b)(a + 4 + b).

a) \(x^2-xy+x-y\)

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

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

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

b) \(2xy-x^2-y^2+16\)

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

\(=4^2-\left(x-y\right)^2\)

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

c) \(x^2-6x-16\)

\(=x^2-6x+9-25\)

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

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

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

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

a)  x² - xy + x - y = x(x - y) + (x - y) = (x - y)(x + 1)

b) 2xy - x² - y² + 16 = 16 - (x - y)²  = (4 - x + y)(4 + x - y)

c) x² - 6x - 16 = (x - 3)² - 25 = (x - 3 - 5)(x - 3 + 5) = (x - 8)(x + 2)

a) \(x^2-xy+x-y\)

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

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

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

b) \(2xy-x^2-y^2+16\)

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

\(=16-\left(x-y\right)^2\)

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

c) \(x^2-6x-16\)

\(=x^2-6x+9-25\)

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

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

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

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

= (x - y)[(x + y)^2 - x]

a) 5x² -20

=  5(x² -4)

=5 (x² -2²)

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

b) 16 - (x+y)²

=4² -(x+y)²

= (4-x-y)(4+x+y) 

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

b, \(16-\left(x+y\right)^2=\left(4-x-y\right)\left(4+x+y\right)\)

mấy bài này áp dụng hđt là được nhé 

5x² - 20 = 5( x² - 4 ) = 5( x - 2 )( x + 2 )

16 - ( x + y )² = 4² - ( x + y ) = ( 4 - x - y )( 4 + x + y ) 

Bài `1`

\(a,5x^2-10xy=5x\left(x-2y\right)\\ b,3x\left(x-y\right)-6\left(x-y\right)=\left(x-y\right)\left(3x-6\right)\\ =3\left(x-y\right)\left(x-2\right)\\ c,2x\left(x-y\right)-4y\left(y-x\right)=2x\left(x-y\right)+4y\left(x-y\right)\\ =\left(x-y\right)\left(2x+4y\right)=2\left(x-y\right)\left(x+2y\right)\\ d,9x^2-9y^2=\left(3x\right)^2-\left(3y\right)^2=\left(3x-3y\right)\left(3x+3y\right)\\ f,xy-xz-y+z=\left(xy-xz\right)-\left(y-z\right)\\ =x\left(y-z\right)-\left(y-z\right)=\left(y-z\right)\left(x-1\right)\)

Bài `3`

\(a,3x^2+8x=0\\ \Leftrightarrow x\left(3x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{8}{3}\end{matrix}\right.\)

\(b,9x^2-25=0\\ \Leftrightarrow\left(3x\right)^2-5^2=0\\ \Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-5=0\\3x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=5\\3x=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

\(c,x^3-16x=0\\ \Leftrightarrow x\left(x^2-16\right)=0\\ \Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

\(d,x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1\in\varnothing\\x=0\end{matrix}\right.\Rightarrow x=0\)

25(x-y)²-16(x+y)²

=[5(x-y)]²-[4(x+y)]²

=[5x-5y]²-[4x+4y]²

=(5x-5y+4x+4y)[(5x-5y)-(4x+4y)]

=(9x-y)(x-9y)

không có số 0 đâu nhá đánh lộn