Violympic toán 8

H24

phân tích đa thức thành nhân tử

a/ \(x^3+x^2+x-3\)

b/ \(x^4+6x^3+7x^2-6x+1\)

c/ \(x^8+x^4+1\)

d/ \(\left(x+y+z\right)^3-x^3-y^3-z^3\)

e/ \(x^3+y^3+z^3-3xyz\)

TP
24 tháng 6 2019 lúc 17:23

a) \(x^3+x^2+x-3\)

\(=x^3-x^2+2x^2-2x+3x-3\)

\(=x^2\left(x-1\right)+2x\left(x-1\right)+3\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+2x+3\right)\)

b) Xét \(x^4+6x^3+7x^2-6x+1=0\)

Dễ thấy x = 0 không thỏa mãn pt

\(x^4+6x^3+7x^2-6x+1\)

\(=x^2\left(x^2+6x+7-\frac{6}{x}+\frac{1}{x^2}\right)\)

\(=x^2\cdot\left(x^2-2+\frac{1}{x^2}+6x-\frac{6}{x}+9\right)\)

\(=x^2\left[\left(x-\frac{1}{x}\right)^2+6\left(x-\frac{1}{x}\right)+9\right]\)

\(=x^2\left(x-\frac{1}{x}+3\right)^2\)

\(=\left[x\left(x-\frac{1}{x}+3\right)\right]^2\)

c) \(x^8+x^4+1\)

\(=x^8+2x^4+1-x^4\)

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

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

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

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

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

d) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=x^3+y^3+z^3+3x^2y+3xy^2+3xz^2+3x^2z+3y^2z+3yz^2+6xyz-x^3-y^3-z^3\)

\(=3\left(x^2y+xy^2+xz^2+x^2z+y^2z+yz^2+2xyz\right)\)

\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

( cái này có kết quả rồi bạn tự phân tích được nhé )

e) \(x^3+y^3+z^3-3xyz\)

\(=x^3+3x^2y+3xy^2+y^3+z^3-3xyz-3x^2y-3xy^2\)

\(=\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-yz-xz-3xy\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)

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