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16x^3y + 0,25yz^3
= 0,25y(64x^3 + z^3 )
= 0,25y(4x^3 + z)(16x^2 - 4x.z + z^2)
\(16x^3y+\frac{1}{4}yz^3\)
\(\text{Phân tích thành nhân tử}\)
\(\frac{y\left(\frac{z}{2}+2x\right)\left(z^2-4xz+16x^2\right)}{2}\)
Câu b sorry nha
a, A =x4 - ( x2 - 2x + 1)
= x4 - ( x - 1)2
= (x2 - x + 1) ( x2 + x - 1)
x3 - x2 + 16x - 16 = x2(x - 1) + 16(x - 1) = (x2 + 16)(x - 1)
x^3-x^2+16x-16 = (x^3-x^2)+(16x-16) = x^2.(x-1) + 16.(x-1) = (x-1)(x^2+16)
1) \(\left(3x^2-3y^2\right)-\left(12x-12y\right)\)
\(=3xy\left(x-y\right)-12\left(x-y\right)\)
\(=\left(3xy-12\right)\left(x-y\right)\)
2) \(4x^3+4xy^2+8x^2y-16x\)
\(=\left(4x^3-16x\right)+\left(4xy^2+8x^2y\right)\)
\(=4x\left(x^2-4\right)+4xy\left(y+2x\right)\)
Ta có : 3x2 - 3y2 - 12x + 12y
= (3x2 - 3y2) - (12x - 12y)
= 3(x2 - y2) - 12(x - y)
= 3(x - y)(x + y) - 4.3.(x - y)
= 3(x - y)(x + y - 4)
\(x^4-4x^3+8x^2-16x+16 \)
\(=x^3\left(x-2\right)-2x^2\left(x-2\right)+4x\left(x-2\right)-8\left(x-2\right)\)
\(=\left(x-2\right)\left(x^3-2x^2+4x-8\right)\)
\(=\left(x-2\right)\left[x^2\left(x-2\right)+4\left(x-2\right)\right]\)
\(=\left(x-2\right)^2\left(x^2+4\right)\)
\(16x^4+8x^2+1-8x^2\)
\(=\left(4x^2+1\right)^2-8x^2\)
\(=\left(4x^2+1-x\sqrt{8}\right)\left(4x^2+1+x\sqrt{8}\right)\)
\(16x^3y+0,25yz^3=16x^3y+\frac{1}{4}yz^3\)
\(=\frac{1}{4}y\left(64x^3+z^3\right)=\frac{y}{4}\left(4x+z\right)\left(16x^2-4xz+z^2\right)\)