Phân tích đa thức thành nhân tử:
a) 9x0 + 24x3y2 + 16y4
b) -y8 + 10y4x3 - 25x6
c) 8x3 + 36x2y + 54xy2 + 27y3
d) -y3 + 12y2x - 48yx2 + 64x3
e) 64x6y4 - 81x2y2
f) 64x6 - 27y6
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a, 2xy^2 ( x^3 -3xy - 4 )
b, x^2 - 4x - 4x +16
= x(x-4) - 4(x-4)
= (x-4) (x-4)
\(a,=8\left(x^3-125\right)=8\left(x-5\right)\left(x^2+5x+25\right)\\ b,=\left(0,1+4x\right)\left(0,01-0,4x+16x^2\right)\\ c,=\left(x+\dfrac{1}{5}y\right)\left(x^2-\dfrac{1}{5}xy+\dfrac{1}{25}y^2\right)\\ d,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ e,=\left(x-1+3\right)\left[\left(x-1\right)^2-3\left(x-1\right)+9\right]\\ =\left(x+2\right)\left(x^2-2x+1-3x+3+9\right)\\ =\left(x+2\right)\left(x^2-5x+13\right)\\ f,=\left(\dfrac{x^2}{2}-y^2\right)\left(\dfrac{x^4}{4}+\dfrac{x^2y^2}{2}+y^4\right)\)
e: \(x^4-2x^3+x^2\)
\(=x^2\cdot x^2-x^2\cdot2x+x^2\cdot1\)
\(=x^2\left(x^2-2x+1\right)\)
\(=x^2\left(x-1\right)^2\)
f: \(27y^3-x^3\)
\(=\left(3y\right)^3-x^3\)
\(=\left(3y-x\right)\left(9y^2+3xy+x^2\right)\)
a) Áp dụng HĐT 5 thu được ( 2 a - 3 b ) 3 .
b) Ta có 8 x 3 + 12 x 2 y + 6 xy 2 + y 3 = ( 2 x + y ) 3 .
Áp dụng HĐT 7 với A = 2x + y; B = z
( 2 x + y ) 3 - z 3 = (2x + y - z)(4 x 2 + y 2 + z 2 + 4xy + 2xz + zy).
Ta có
8 x 3 + 12 x 2 y + 6 x y 2 + y 3 = ( 2 x ) 3 + 3 . ( 2 x ) 2 y + 3 . 2 x . y 2 + y 3 = ( 2 x + y ) 3
Đáp án cần chọn là: B
\(a,=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\\ b,=4x^2\left(x^2+2x+1\right)=4x^2\left(x+1\right)^2\\ c,=xy^2\left(x^2-2xy+y^2\right)=xy^2\left(x-y\right)^2\\ d,=\left(x-y\right)\left(x+y\right)-7\left(x-y\right)=\left(x-y\right)\left(x+y-7\right)\\ e,=\left(5x-2y\right)\left(5x+2y\right)\\ f,=x^2+3x+4x+12=\left(x+3\right)\left(x+4\right)\\ i,=x^2+2x-7x-14=\left(x+2\right)\left(x-7\right)\)
\(8x^3+12x^2y+6xy^2+y^3-z^3\)
\(=\left(2x+y\right)^3-z^3\)
\(=\left(2x+y-z\right)\left[4x^2+z\left(2x+y\right)+z^2\right]\)
a, 8a3 - 36a2 +54ab2 - 27b3
=(8a3-36a2b +54ab2 - 27b3)
=(2a-3b)2
=(2a-3b)(2a-3b)(2a-3b)
b, 8x3 + 12x2y + 6xy2 + y3 - z 3
=(8x3 + 12x2y + 6xy2 + y3) - z3
=(2x + y)3 - y3
=(2x + y +z) . [ (2x + Y)2 + 2(2x + y)+ z2
= (2x + y + z)(4x2 + 4xy + y2 + 4x + 2y + z2
b) \(-y^8+10y^4x^3-25x^6\)
\(=-\left(y^8-10y^4x^3+25x^6\right)\)
\(=-\left[\left(y^4\right)^2-2.y^4.5x^3+\left(5x^3\right)^2\right]\)
\(=-\left(y^4-5x^3\right)^2\)
c) \(8x^3+36x^2y+54xy^2+27y^3\)
\(=\left(2x\right)^3+3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2+\left(3y\right)^3\)
\(=\left(2x+3y\right)^3\)
d) \(-y^3+12y^2x-48yx^2+64x^3\)
\(=-\left(y^3-12y^2x+48yx^2-64x^3\right)\)
\(=-\left[y^3-3.y^2.4x+3.y.\left(4x\right)^2-\left(4x\right)^3\right]\)
\(=-\left(y-4x\right)^3\)
e) \(64x^6y^4-81x^2y^2\)
\(=\left(8x^3y^2\right)^2-\left(9xy\right)^2\)
\(=\left(8x^3y^2-9xy\right)\left(8x^3y^2+9xy\right)\)
f) \(64x^6-27y^6\)
\(=\left(4x^2\right)^3-\left(3y^2\right)^3\)
\(=\left(4x^2-3y^2\right)\left[\left(4x^2\right)^2+4x^2.3y^2+\left(3y^2\right)^2\right]\)
\(=\left(4x^2-3y^2\right)\left(16x^4+12x^2y^2+9x^4\right)\)