Câu hỏi:

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Phân tích mỗi đa thức sau thành nhân tử:

a) 8x3yz+12x2yz+6xyz+yz8{x^3}yz + 12{x^2}yz + 6xyz + yz;        

b) 81x4(z2y2)z2+y281{x^4}\left( {{z^2} - {y^2}} \right) - {z^2} + {y^2};

c) x38y327+x2y3\frac{{{x^3}}}{8} - \frac{{{y^3}}}{{27}} + \frac{x}{2} - \frac{y}{3};                                                    

d) x6+x4+x2y2+y4y6{x^6} + {x^4} + {x^2}{y^2} + {y^4} - {y^6}.

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Lời giải

Hướng dẫn giải:

Lời giải:

a) 8x3yz+12x2yz+6xyz+yz8{x^3}yz + 12{x^2}yz + 6xyz + yz

=yz(8x3+12x2+6x+1) = yz\left( {8{x^3} + 12{x^2} + 6x + 1} \right)

=yz[(2x)3+3.(2x)2.1+3.2x.12+13] = yz\left[ {{{\left( {2x} \right)}^3} + 3.{{\left( {2x} \right)}^2}.1 + 3.2x{{.1}^2} + {1^3}} \right]

=yz(2x+1)3 = yz{\left( {2x + 1} \right)^3}.

b) 81x4(z2y2)z2+y281{x^4}\left( {{z^2} - {y^2}} \right) - {z^2} + {y^2}

=81x4(z2y2)(z2y2) = 81{x^4}\left( {{z^2} - {y^2}} \right) - \left( {{z^2} - {y^2}} \right)

=(z2y2)(81x41) = \left( {{z^2} - {y^2}} \right)\left( {81{x^4} - 1} \right)

=(zy)(z+y)[(9x2)212] = \left( {z - y} \right)\left( {z + y} \right)\left[ {{{\left( {9{x^2}} \right)}^2} - {1^2}} \right]

=(zy)(z+y)(9x2+1)(9x21) = \left( {z - y} \right)\left( {z + y} \right)\left( {9{x^2} + 1} \right)\left( {9{x^2} - 1} \right)

=(zy)(z+y)(9x2+1)[(3x)212] = \left( {z - y} \right)\left( {z + y} \right)\left( {9{x^2} + 1} \right)\left[ {{{\left( {3x} \right)}^2} - {1^2}} \right]

=(zy)(z+y)(9x2+1)(3x+1)(3x1) = \left( {z - y} \right)\left( {z + y} \right)\left( {9{x^2} + 1} \right)\left( {3x + 1} \right)\left( {3x - 1} \right).

c) x38y327+x2y3\frac{{{x^3}}}{8} - \frac{{{y^3}}}{{27}} + \frac{x}{2} - \frac{y}{3}

=(x38y327)+(x2y3) = \left( {\frac{{{x^3}}}{8} - \frac{{{y^3}}}{{27}}} \right) + \left( {\frac{x}{2} - \frac{y}{3}} \right)

=[(x2)3(y3)3]+(x2y3) = \left[ {{{\left( {\frac{x}{2}} \right)}^3} - {{\left( {\frac{y}{3}} \right)}^3}} \right] + \left( {\frac{x}{2} - \frac{y}{3}} \right)

=(x2y3)[(x2)2+x2.y3+(y3)2]+(x2y3) = \left( {\frac{x}{2} - \frac{y}{3}} \right)\left[ {{{\left( {\frac{x}{2}} \right)}^2} + \frac{x}{2}.\frac{y}{3} + {{\left( {\frac{y}{3}} \right)}^2}} \right] + \left( {\frac{x}{2} - \frac{y}{3}} \right)

=(x2y3).(x24+xy6+y29+1) = \left( {\frac{x}{2} - \frac{y}{3}} \right).\left( {\frac{{{x^2}}}{4} + \frac{{xy}}{6} + \frac{{{y^2}}}{9} + 1} \right).

d) x6+x4+x2y2+y4y6{x^6} + {x^4} + {x^2}{y^2} + {y^4} - {y^6}

=(x6y6)+(x4+2x2y2+y4)x2y2 = \left( {{x^6} - {y^6}} \right) + \left( {{x^4} + 2{x^2}{y^2} + {y^4}} \right) - {x^2}{y^2}

=[(x3)2(y3)2]+[(x2)2+2x2y2+(y2)2](xy)2 = \left[ {{{\left( {{x^3}} \right)}^2} - {{\left( {{y^3}} \right)}^2}} \right] + \left[ {{{\left( {{x^2}} \right)}^2} + 2{x^2}{y^2} + {{\left( {{y^2}} \right)}^2}} \right] - {\left( {xy} \right)^2}

=(x3y3)(x3+y3)+[(x2+y2)2(xy)2] = \left( {{x^3} - {y^3}} \right)\left( {{x^3} + {y^3}} \right) + \left[ {{{\left( {{x^2} + {y^2}} \right)}^2} - {{\left( {xy} \right)}^2}} \right]

=(xy)(x2+xy+y2)(x+y)(x2xy+y2)+(x2+y2xy)(x2+y2+xy) = \left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right) + \left( {{x^2} + {y^2} - xy} \right)\left( {{x^2} + {y^2} + xy} \right)

=(x2+y2+xy)(x2+y2xy)[(xy)(x+y)+1] = \left( {{x^2} + {y^2} + xy} \right)\left( {{x^2} + {y^2} - xy} \right)\left[ {\left( {x - y} \right)\left( {x + y} \right) + 1} \right]

=(x2+y2+xy)(x2+y2xy)(x2y2+1) = \left( {{x^2} + {y^2} + xy} \right)\left( {{x^2} + {y^2} - xy} \right)\left( {{x^2} - {y^2} + 1} \right).

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