Phân tích đa thức thành nhân tử:
\(x^6-64y^{12}\)
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\(=\left(2x\right)^3-\left(4y^2\right)^3\)
Sau đó thì sử dụng HĐT số 7
Có:\(x^4+64y^4\)
\(=\left(x^4+16x^2y^2+64y^4\right)-16x^2y^2\)
\(=\left(x^2+8y^2\right)^2-\left(4xy\right)^2\)
\(=\left(x^2+4xy+8y^2\right)\left(x^2-4xy+8y^2\right)\)
Linz
= 64y4 + 32xy3 + 8y2x2 - 32xy3 -16x2y2 - 4x3y + 8x2y2 +4x3y +x4
= 8y2 ( 8y2 + 4xy + x2 ) - 4xy ( 8y2 + 4xy + x2 ) + x2 ( 8y2 + 4xy + x2 )
= ( 8y2 - 4xy + x2 ) ( 8y2 + 4xy + x2 )
a,\(0,04x^2-64y^2=\left(0,2x\right)^2-\left(8y\right)^2=\left(0,2x-8y\right)\left(0,2x+8y\right)\)
b,\(-x^3+9x^2-27x+27=-x^3+3x^2+6x^2-18x-9x+27\)
\(=-x^2\left(x-3\right)+6x\left(x-3\right)-9\left(x-3\right)=\left(-x^2+6x-9\right)\left(x-3\right)\)
\(=-\left(x-3\right)^3\)
Nhớ tick mình nha bạn,cảm ơn nhiều nha.
a: \(15x^2-5x^3=5x^2\left(3-x\right)\)
b: \(8x^3-y^3+4x^2y-2xy^2\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)+2xy\left(2x-y\right)\)
\(=\left(2x-y\right)\left(4x^2+4xy+y^2\right)\)
\(=\left(2x-y\right)\left(2x+y\right)^2\)
c: Ta có: \(x^8+64y^4\)
\(=x^8+16x^4y^2+64y^4-16x^4y^2\)
\(=\left(x^4+8y^2\right)^2-\left(4x^2y\right)^2\)
\(=\left(x^2-4x^2y+8y^2\right)\left(x^2+4x^2y+8y^2\right)\)
\(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)-3x^2\)
\(=4\left[\left(x+5\right)\left(x+12\right)\right]\left[\left(x+6\right)\left(x+10\right)\right]-3x^2\)
\(=4\left(x^2+17x+60\right)\left(x^2+16x+60\right)-3x^2\)
\(=\left(2x^2+34x+120\right)\left(2x^2+32x+60\right)-3x^2\)
\(=\left(2x^2+33x+120\right)^2-x^2-3x^2\)
\(=\left(2x^2+33x+120-2x\right)\left(2x^2+33x+120+2x\right)\)
\(=\left(2x+15\right)\left(x+8\right)\left(2x^2+35x+120\right)\)
a, 8x^3 - 1/8
= (2x)^3 - (1/2)^3
= ( 2x - 1/2) ( 4x^2 + x + 1/4)
b, 1/25.x^2 - 64y^2 = (1/5x)^2 - (8y)^2 = ( 1/5x - 8y)(1/5x+8y)
Đúng xho mình nha )
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)
\(x^6-64x^{12}=\left(x^3\right)^2-\left(8x^6\right)^2=\left(x^3-8x^6\right)\left(x^3+8x^6\right).\)
\(=x^6\left(1-8x^3\right)\left(1+8x^3\right)=x^6\left(1-2x\right)\left(1+2x+4x^2\right)\left(1+2x\right)\left(1-2x+4x^2\right)\)