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a) \(4x^3y-12x^2y^3-8x^4y^3\)
\(=4x^2y\left(x-3y^2-2x^2y^2\right)\)
b) \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x-y+1\right)\left(x+y+1\right)\)
c) \(x^3-2x^2+x-xy^2\)
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left(x-y-1\right)\left(x+y-1\right)\)
d) \(x\left(x-2y\right)+3\left(2y-x\right)\)
\(=x\left(x-2y\right)-3\left(x-2y\right)\)
\(=\left(x-3\right)\left(x-2y\right)\)
e) \(x^2+4\)
\(=\left(x^4+4x^2+4\right)-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
f) \(5x^2-7x-6\)
\(=\left(5x^2-10x\right)+\left(3x-6\right)\)
\(=5x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(5x+3\right)\left(x-2\right)\)
Đây là cách hiện đại :
\(x^4-2x^3+2x-1\)
\(=\left(x^4-1\right)-\left(2x^3-2x\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(\left(x^2+1\right)-2x\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(\left(x^2+1\right)-2x\right)\)
a,=\(x^4-x^3-x^3+x^2-x^2+x+x-1\)
cu hai so nhom 1 nhom roi dat thua so chung la xong
b,x^4+x^3+x^3+x^2+x^2+x+x+1
cu hai so lai nhom 1 nhom va dat thua so chung
a) x^2 - 4 + ( x - 2 )^2
= ( x- 2 )(x + 2 ) + ( x- 2)^2
= ( x - 2 ) ( x + 2 + x - 2 )
= 2x (x-2)
b) x^3 - 2x^2 + x - xy^2
= x ( x^2 - 2x + 1 - y^2)
= x [ ( x - 1 )^2 - y^2 ]
= x(x - 1 - y)( x - 1 + y )
c) x^3 - 4x^2 - 12x + 27
= x^3 + 3x^2 - 7x^2 - 21x + 9x + 27
= x^2 ( x + 3 ) - 7x ( x+ 3 ) + 9(x + 3 )
Để hai lần nha
= ( x+ 3 )(x^2 - 7x + 9 )
\(x^2-4+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2+x-2\right)\)
\(=2x\left(x-2\right)\)
hk tốt
^^
a) 3 ( 4x +5y )
b) (x-y) (x+y+1)
d) (x3)2 - 82 = (x3- 8) (x3+8)
a, 12x+15y = 3(4x+5y)
b, x^2 - y^2 - x - y = (x - y)(x + y) - (x + y) = (x + y)(x - y - 1)
c, x^3 + 2x^2 + x = x(x^2 + 2x + 1) = x(x + 1)^2
d, x^6 - 64 = (x^3 - 8)(x^3 + 8) = (x - 2)(x^2 + 2x + 4)(x + 2)(x^2 - 2x + 4)
a) \(=x\left(x^2-2xy+y^2\right)=x\left(x-y\right)^2\)
b) \(=\left(x^2+2x\right)+\left(10x+20\right)=x\left(x+2\right)+10\left(x+2\right)=\left(x+2\right)\left(x+10\right)\)
c) đặt \(x^2+x+1=t\)
\(\left(x^2+x+1\right)\left(x^2+x+4\right)+2=t\left(t+3\right)+2=t^2+3t+2=\left(t^2+t\right)+\left(2t+2\right)=t\left(t+1\right)+2\left(t+1\right)=\left(t+1\right)\left(t+2\right)=\left(x^2+x+2\right)\left(x^2+x+3\right)\)