Violympic toán 8

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ITACHY

Phân tích thành nhân tử:

a, a2b2(a-b)+b2c2(b-c)+c2a2(c-a)

b, x2-4xy+4y2-7x+14y+6

c, (x2+6x+8)(x2+8x+15)-24

Yukru
7 tháng 8 2018 lúc 17:54

a) \(a^2b^2\left(a-b\right)+b^2c^2\left(b-c\right)+c^2a^2\left(c-a\right)\)

\(=a^3b^2-a^2b^3+b^2c^2\left(b-c\right)+c^3a^2-c^2a^3\)

\(=\left(a^3b^2-c^2a^3\right)-\left(a^2b^3-c^3a^2\right)+b^2c^2\left(b-c\right)\)

\(=a^3\left(b^2-c^2\right)-a^2\left(b^3-c^3\right)+b^2c^2\left(b-c\right)\)

\(=a^3\left(b-c\right)\left(b+c\right)-a^2\left(b-c\right)\left(b^2+bc+c^2\right)+b^2c^2\left(b-c\right)\)

\(=\left(b-c\right)\left[a^3\left(b+c\right)-a^2\left(b^2+bc+c^2\right)+b^2c^2\right]\)

\(=\left(b-c\right)\left[a^3\left(b+c\right)-a^2b^2-a^2bc-a^2c^2+b^2c^2\right]\)

\(=\left(b-c\right)\left[a^3\left(b+c\right)-a^2b\left(b+c\right)-c^2\left(a^2-b^2\right)\right]\)

\(=\left(b-c\right)\left[\left(b+c\right)\left(a^3-a^2b\right)-c^2\left(a-b\right)\left(a+b\right)\right]\)

\(=\left(b-c\right)\left[\left(b+c\right)a^2\left(a-b\right)-c^2\left(a-b\right)\left(a+b\right)\right]\)

\(=\left(b-c\right)\left(a-b\right)\left[\left(b+c\right)a^2-c^2\left(a+b\right)\right]\)

\(=\left(b-c\right)\left(a-b\right)\left(a^2b+a^2c-c^2a-c^2b\right)\)

\(=\left(b-c\right)\left(a-b\right)\left[\left(a^2b-c^2b\right)+\left(a^2c-c^2a\right)\right]\)

\(=\left(b-c\right)\left(a-b\right)\left[b\left(a^2-c^2\right)+ac\left(a-c\right)\right]\)

\(=\left(b-c\right)\left(a-b\right)\left[b\left(a-c\right)\left(a+c\right)+ac\left(a-c\right)\right]\)

\(=\left(b-c\right)\left(a-b\right)\left(a-c\right)\left[b\left(a+c\right)+ac\right]\)

\(=\left(b-c\right)\left(a-b\right)\left(a-c\right)\left(ab+bc+ac\right)\)

b) \(x^2-4xy+4y^2-7x+14y+6\)

\(=x^2-2.x.2y+\left(2y\right)^2-7\left(x-2y\right)+6\)

\(=\left(x-2y\right)^2-7\left(x-2y\right)+6\)

\(=\left(x-2y\right)^2-\left(x-2y\right)-6\left(x-2y\right)+6\)

\(=\left(x-2y\right)\left(x-2y-1\right)-6\left(x-2y-1\right)\)

\(=\left(x-2y-1\right)\left(x-2y-6\right)\)

c) \(\left(x^2+6x+8\right)\left(x^2+8x+15\right)-24\)

\(=\left(x+2\right)\left(x+4\right)\left(x+3\right)\left(x+5\right)-24\)

\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+4\right)\left(x+3\right)\right]-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt x2 + 7x + 11 = a, ta được:

\(=\left(a-1\right)\left(a+1\right)-24\)

\(=a^2-1-24\)

\(=a^2-25\)

\(=\left(a-5\right)\left(a+5\right)\)

\(=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)


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