Violympic toán 8

LT

Phân tích đa thức thành nhân tử:

a. \(x^6-x^4-9x^3+9x^2\)

b. \(x^4+x^3+6x^2+5x+5\)

c. \(\left(12x^2-12xy+3y^2\right)-10\left(2x-y\right)+8\)

BK
1 tháng 8 2018 lúc 20:41

\(x^6-x^4-9x^3+9x^2\)

\(=x^4\left(x^2-1\right)-9x^2\left(x-1\right)\)

\(=x^4\left(x+1\right)\left(x-1\right)-9x^2\left(x-1\right)\)

\(=\left(x^5+x^4\right)\left(x-1\right)-9x^2\left(x-1\right)\)

\(=\left(x-1\right)\left(x^5+x^4-9x^2\right)\)

\(x^4+x^3+6x^2+5x+5\)

\(=x^4+x^3+x^2+5x^2+5x+5\)

\(=x^2\left(x^2+x+1\right)+5\left(x^2+x+1\right)\)

\(=\left(x^2+5\right)\left(x^2+x+1\right)\)

\(\left(12x^2-12xy+3y^2\right)-10\left(2x-y\right)+8\)

\(=3\left(4x^2-4xy+y^2\right)-10\left(2x-y\right)+8\)

\(=3\left[\left(2x\right)^2-2.2x.y+y^2\right]-10\left(2x-y\right)+8\)

\(=3\left(2x-y\right)^2-10\left(2x-y\right)+8\)

\(=\left(2x-y\right)\left[3\left(2x-y\right)-10\right]+8\)

\(=\left(2x-y\right)\left(6x-3y-10\right)+8\)

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