[Đại 8] Phân tích đa thức thành nhân tử

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0973573959thuy

$a) y(x - 2z)^2 + 8xyz + x(y - 2z)^2 - 2z(x + y)^2$

$= y(x^2 - 4xz + 4z^2) + 8xyz + x(y^2 - 4yz + 4z^2) - 2z(x^2 + y^2 + 2xy)$

$= (yx^2 + xy^2) + (4yz^2 + 4xz^2) - (2zx^2 + 2zy^2 + 4xyz)$

$= xy(x + y) + 4z^2(x + y) - 2z(x^2 + y^2 + 2xy)$

$= xy(x + y) + 4z^2(x + y) - 2z(x + y)^2$

$= (x + y)(xy + 4z^2 - 2xz - 2yz)$

$= (x + y)[y(x - 2z) - 2z(x - 2z)]$

$= (x + y)(y - 2z)(x - 2z)$

b) $[(x^2 + y^2)(a^2 + b^2) + 4abxy]^2 - 4[xy(a^2 + b^2) + ab(x^2 + y^2)]^2$

$= (x^2a^2 + x^2b^2 + y^2a^2 + y^2b^2 + 4abxy)^2 - 4(xya^2 + xyb^2 + abx^2 + aby^2)^2$

$= [(a^2y^2 + b^2x^2 + 2abxy) + (a^2x^2 + b^2y^2 + 2abxy)]^2 - 4[ax(ay + bx) + by(ay + bx)]^2$

$= [(ay + bx)^2 + (ax + by)^2]^2 - 4[(ax+by)(ay + bx)]^2$

$= (ay + bx)^4 + (ax + by)^4 + 2(ay + bx)^2(bx + ay)^2 - 4(ax + by)^2(ay + bx)^2$

$= (ay + bx)^4 + (ax + by)^4 - 2(ay + bx)^2(bx + ay)^2$

$= [(ay + bx)^2 - (ax + by)^2]^2$

$= (a^2y^2 + b^2x^2 + 2abxy - a^2x^2 - b^2y^2 - 2abxy)^2$

$= [y^2(a^2 - b^2) - x^2(a^2 - b^2)]^2$

$= (a - b)^2(a + b)^2(y - x)^2(x + y)^2$

P.s : Gõ Latex nhiều, nhầm chỗ nào thì thông cảm nhá :))
 
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