Question

[Trang Nghiêm]

bài 4:phân tích mỗi đa thức sau thành nhân tích

a, 8³ yz + 12²yz + 6xyz + yz

b,81x⁴(z² - y²) - z² + y²

c,\(\dfrac{x^3}{8}\) - \(\dfrac{y^3}{27}\) +\(\dfrac{x}{2}\) - \(\dfrac{y}{3}\) 

d, x⁶ + x⁴ + x² y² + y⁴ - y⁶

Answer 1

[Toru]

a, \(8^3yz+12^2yz+6xyz+yz\)

\(=512yz+144yz+6xyz+yz\)

\(=yz\left(512+14+6x+1\right)\)

\(=yz\left(527+6x\right)\)

$---$

b, \(81x^4\left(z^2-y^2\right)-z^2+y^2\)

\(=81x^4\left(z^2-y^2\right)-\left(z^2-y^2\right)\)

\(=\left(z^2-y^2\right)\left(81x^4-1\right)\)

\(=\left(z-y\right)\left(z+y\right)\left[\left(9x^2\right)^2-1^2\right]\)

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

\(=\left(z-y\right)\left(z+y\right)\left[\left(3x\right)^2-1^2\right]\left(9x^2+1\right)\)

\(=\left(z-y\right)\left(z+y\right)\left(3x-1\right)\left(3x+1\right)\left(9x^2+1\right)\)

$---$

c, \(\dfrac{x^3}{8}-\dfrac{y^3}{27}+\dfrac{x}{2}-\dfrac{y}{3}\)

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

\(=\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{4}+\dfrac{xy}{6}+\dfrac{y^2}{9}\right)+\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\)

\(=\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{4}+\dfrac{xy}{6}+\dfrac{y^2}{9}+1\right)\)

$---$

d, \(x^6+x^4+x^2y^2+y^4-y^6\)

\(=\left(x^6-y^6\right)+\left(x^4+x^2y^2+y^4\right)\)

\(=\left[\left(x^2\right)^3-\left(y^2\right)^3\right]+\left(x^4+x^2y^2+y^4\right)\)

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

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

$Toru$