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\(1,4x^4+4x^2y^2-8y^4\)
\(=4\left(x^4+x^2y^2-y^4-y^4\right)\)
\(=4\left[\left(x^4-y^4\right)+\left(x^2y^2-y^4\right)\right]\)
\(=4\left[\left(x^2+y^2\right)\left(x^2-y^2\right)+y^2\left(x^2-y^2\right)\right]\)
\(=4\left(x^2-y^2\right)\left(x^2+y^2+y^2\right)\)
\(=4\left(x-y\right)\left(x+y\right)\left(x^2+2y^2\right)\)
\(2,12x^2y-18xy^2-30y^3\)
\(=6y\left(2x^2-3xy-5y^2\right)\)
\(=6y\left[\left(2x^2+2xy\right)-\left(5xy+5y^2\right)\right]\)
\(=6y\left[2x\left(x+y\right)-5y\left(x+y\right)\right]\)
\(=6y\left(x+y\right)\left(2x-5y\right)\)
a)
\(x^5+x^3-x^2-1\)
\(=x^3\left(x^2+1\right)-\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^3-1\right)\)
\(=\left(x^2+1\right)\left(x-1\right)\left(x^2+x+1\right)\)
b)
\(x^2-3x^3-x+3\)
\(=x\left(x-1\right)-3\left(x^3-1\right)\)
\(=x\left(x-1\right)-3\left(x-1\right)\left(x^2+x+1\right)\)
\(=\left(x-1\right)\left(x-3x^2-3x-3\right)\)
\(=\left(x-1\right)\left(-3x^2-2x-3\right)\)
c)
\(x^2-6x+8\)
\(=x^2-2.x.3+9-1\)
\(=\left(x-3\right)^2-1\)
\(=\left(x-3-1\right)\left(x-3+1\right)\)
\(=\left(x-4\right)\left(x-2\right)\)
d)
\(4x^4-4x^2y^2-8y^4\)
\(=\left(2x^2\right)^2-2.\left(2x^2\right)y^2+y^2-9y^4\)
\(=\left(2x^2-y\right)^2-\left(3y^2\right)^2\)
\(=\left(2x^2-y-3y^2\right)\left(2x^2-y+3y^2\right)\)
a,nhóm x*5 với x*3,x*2 và 1: (x*5+ x*3) - (x*2+1) =x*3.(x*2+1)-(x*2+1) =....., câu b nhóm x*2 và -3x*3,x và 3, câu c bang (x-3)*2-1 =..., câu d đat 4 ra.
a) \(x^5+x^3-x^2-1=x^3\left(x^2+1\right)-\left(x^2+1\right)=\left(x^3-1\right)\left(x^2+1\right)\)
b) \(x^2-3x^3-x+3=x\left(x-1\right)-3\left(x^3-1\right)=x\left(x-1\right)-3\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left[\left(x-3\right)-\left(x^2+x+1\right)\right]=\left(x-1\right)\left(-x^2-4\right)\)c) \(x^2-6x+8=x^2-6x+9-1=\left(x-3\right)^2-1=\left(x-2\right)\left(x-4\right)\)
d) \(4x^4+4x^2y^2-8y^4=4x^4+4x^2y^2+y^4-9y^4=\left(2x^2+y^2\right)^2-9y^4=\left(2x^2+4y^2\right)\left(2x^2-2y^2\right)=2\left(x^2+2y^2\right)2\left(x^2-y^2\right)=4\left(x^2+2y^2\right)\left(x+y\right)\left(x-y\right)\)
a) x^4 - x^3 - x + 1
= x^3 ( x - 1 ) - ( x- 1 )
= ( x^3 - 1 )(x - 1)
= ( x- 1 )^2 (x^2 + x + 1 )
a)x4-x3-x+1
=x3(x-1)-(x-1)
=(x-1)(x3-1)
=(x-1)(x-1)(x2+x+1)
=(x-1)2(x2+x+1)
b)5x2-4x+20xy-8y
(sai đề)
\(A=\left(x^2+x\right)^2-14\left(x^2+x\right)+24\)
Đặt \(x^2+x=t\), ta có:
\(A=t^2-14t+24\)
\(=t^2-2t-12t+24\)
\(=t\left(t-2\right)-12\left(t-2\right)\)
\(=\left(t-2\right)\left(t-12\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x-12\right)\)
\(B=\left(x^2+x\right)^2+4x^2+4x-12\)
\(=\left(x^2+x\right)^2+4\left(x^2+x\right)-12\)
Đặt \(x^2+x=t\), ta có:
\(B=t^2+4t-12\)
\(=t^2+6t-2t-12\)
\(=t\left(t+6\right)-2\left(t+6\right)\)
\(=\left(t+6\right)\left(t-2\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(C=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)
Đặt \(x^2+5x+4=t\), ta có:
\(C=t\left(t+2\right)+1\)
\(=t^2+2t+1\)
\(=\left(t+1\right)^2\)
\(=\left(x^2+5x+4+1\right)^2\)
\(=\left(x^2+5x+5\right)^2\)
\(D=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(x^2+8x+7=t\), ta có:
\(D=t\left(t+8\right)+15\)
\(=t^2+8t+15\)
\(=t^2+3t+5t+15\)
\(=t\left(t+3\right)+5\left(t+3\right)\)
\(=\left(t+3\right)\left(t+5\right)\)
\(=\left(x^2+8x+7+3\right)\left(x^2+8x+7+5\right)\)
\(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)
\(F=\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
Đặt \(x^2+x+1=t\), ta có:
\(F=t\left(t+1\right)-12\)
\(=t^2+t-12\)
\(=t^2+4t-3t-12\)
\(=t\left(t+4\right)-3\left(t+4\right)\)
\(=\left(t+4\right)\left(t-3\right)\)
\(=\left(x^2+x+1+4\right)\left(x^2+x+1-3\right)\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(E=x^4+2x^3+5x^2+4x-12\)
\(=x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12\)
\(=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+3x^2+8x+12\right)\)
\(=\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)\)
\(=\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
\(a,4x^4-8x^3+4x^2\)
\(=4x^2\cdot\left(x^2-2x+1\right)\)
\(=4x^2\cdot\left(x-1\right)^2\)
\(b,x^2-y^2+5\cdot\left(y-x\right)\)
\(=\left(x-y\right)\cdot\left(x+y\right)-5\cdot\left(x-y\right)\)
\(=\left(x-y\right)\cdot\left(x+y-5\right)\)
\(c,3x^2-6xy+3y^2-12z^2\)
\(=3\cdot\left(x^2-2xy+y^2-4x^2\right)\)
\(=3\cdot\left[\left(x-y\right)^2-\left(2x\right)^2\right]\)
\(=3\cdot\left(x-y-2x\right)\cdot\left(x-y+2x\right)\)
a)\(x^2+4x-4y^2-8y\)
\(=x^2+2xy+4x-2xy-4y^2-8y\)
\(=x\left(x+2y+4\right)-2y\left(x+2y+4\right)\)
\(=\left(x-2y\right)\left(x+2y+4\right)\)
b)sai đề
c)sai đề tiếp
a)x2+4x-4y2-8y=(x2-4y2)+(4x-8y)
=(x+2y(x-2y)+4(x-2y)
=(x-2y)(x+2y+4)
\(x^2-y^2+4x+4\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(4x^2-y^2+8\left(y-2\right)\)
\(=4x^2-\left(y^2-8y+16\right)\)
\(=4x^2-\left(y-4\right)^2\)
\(=\left(2x+y-4\right)\left(2x-y+4\right)\)
\(\text{a) }x^5+x^3-x^2-1\\ \\=\left(x^5-x^2\right)+\left(x^3-1\right)\\ \\=x^2\left(x^3-1\right)+\left(x^3-1\right)\\ \\=\left(x^2+1\right)\left(x^3-1\right)\\ \\=\left(x^2+1\right)\left(x-1\right)\left(x^2+x+1\right)\\ \)
\(\text{b) }x^2-x-12\\ \\=x^2-4x+3x-12\\ \\ =\left(x^2-4x\right)+\left(3x-12\right)\\ =x\left(x-4\right)+3\left(x-4\right)\\ \\=\left(x-4\right)\left(x+3\right)\\ \)
a) x5 + x3 - x2 - 1 = ( x5 + x3 ) - ( x2 + 1)
= x3 . ( x2 + 1 ) - ( x2 + 1 )
= ( x2 + 1 ) . ( x3 - 1 )
= ( x2 + 1 ) . ( x - 1 ) . ( x2 + x + 1 )
b) x2 - x - 12 = x2 - 4x + 3x - 12
= x . ( x - 4 ) + 3 . ( x - 4 )
= ( x - 4 ) . ( x + 3 )
c) 4x4 + 4x2y2 - 8y4 = 4x4 - 4x2y2 + 8x2y2 - 8y4
= 4x2 . ( x2 - y2 ) + 8y2 . ( x2 - y2 )
= ( x2 - y2 ) . ( 4x2 + 8y2 )
= 4 . ( x - y ) . ( x + y ) . ( x2 + 2y2 )