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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) \(x^3-xy+2y-8=\left(x-2\right)\left(x^2+2x+4\right)-y\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+4-y\right)\)
\(2\left(x-1\right)^3-5\left(x-1\right)^2-\left(x-1\right)=\left(x-1\right)\left[2\left(x-1\right)^2-5\left(x-1\right)-1\right]=\left(x-1\right)\left(2\left(x^2-2x+1\right)-5x+5-1\right)=\left(x-1\right)\left(2x^2-4x+2-5x+5-1\right)=\left(x-1\right)\left(2x^2-9x+6\right)\)
\(2\left(x-1\right)^3-5\left(x-1\right)^2-\left(x-1\right)\)
\(=\left(x-1\right)\left[2\left(x-1\right)^2-5\left(x-1\right)-1\right]\)
\(=\left(x-1\right)\left[2\left(x^2-2x+1\right)-5\left(x-1\right)-1\right]\)
\(=\left(x-1\right)\left(2x^2-4x+2-5x+5-1\right)\)
\(=\left(x-1\right)\left(2x^2-9x+6\right)\)
a) x3-2x2-x+2
=x(x2-1)+2(-x2+1)
=x(x2-1)-2(x2-1)
=(x2-1)(x-2)
b)
x2+6x-y2+9
=x2+6x+9-y2
=(x+3)2-y2
=(x+3-y)(x+3+y)
\(1,x^2-y^2+4x-4y\)
\(\left(x-y\right)\left(x+y\right)+4\left(x-y\right)\)
\(\left(x-y\right)\left(x+y+4\right)\)
\(x^2+2x-4y^2-4y\)
\(\left(x-2y\right)\left(x+2y\right)+2\left(x-2y\right)\)
\(\left(x-2y\right)\left(x+2y+2\right)\)
\(3,3x^2-4y+4x-3y^2\)
\(3\left(x^2-y^2\right)-4\left(x-y\right)\)
\(3\left(x-y\right)\left(x+y\right)-4\left(x-y\right)\)
\(\left(x-y\right)\left(3x+3y-4\right)\)
\(x^4-6x^3+54x-81\)
\(x^4+3x^3-9x^3+27x^2-27x^2+81x-27x-81\)
\(\left(x^4+3x^3\right)-\left(9x^3+27x^2\right)+\left(27x^2+81x\right)-\left(27x+81\right)\)
\(x^3\left(x+3\right)-9x^2\left(x+3\right)+27x\left(x+3\right)-27\left(x+3\right)\)
\(\left(x+3\right)\left(x^3-9x^2+27x-27\right)\)
\(\left(x+3\right)\left(x-3\right)^3\)
a, x2+2xy+y2+2x+2y-15
<=> (x+y )2+2(x+y)+1-16
Đặt x+y =a
<=> a2+2a+1-42
<=> (a+1)2-42
<=> (a+5)(a-3) =>( x+y+5)(x+y-3)
b, x2-4xy+4y2-2x-4y-35
<=> (x-2y)2-2(x-2y)+1-36
Đặt (x-2y) =b
=> b2-2b+1-62
<=> (b-1)2-62
<=> (b-7)(b+5)=> (x-2y-7)(x-2y+5)
c,
a,A= x^2+2xy+y^2+2x+2y-15
= (x+y)^2+(x+y)-15
Đặt x+y=a, ta có:
A=a^2+2a-15
=a^2+2a+1-16
=(a+1)^2-4^2
=(a+1+4)(a+1-4)
=(a+5)(a-3)
Thay a=x+y, ta có: A=(x+y+5)(x+y-3).
\(x^2+6x-7\)
\(=x^2+7x-x-7\)
\(=\left(x^2+7x\right)-\left(x+7\right)\)
\(=x\left(x+7\right)-\left(x+7\right)\)
\(=\left(x-1\right)\left(x+7\right)\)
\(x^2+4x^2\) ( bạn xem đề bài đi chứ câu này ko có nhân tử chung)