Câu hỏi:

89 lượt xem
Tự luận

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}.

Xem đáp án

Lời giải

Hướng dẫn giải:

Lời giải:

a) \(8{x^3}yz + 12{x^2}yz + 6xyz + yz\)

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

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

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

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

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

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

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

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

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

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

c) \[\frac{{{x^3}}}{8} - \frac{{{y^3}}}{{27}} + \frac{x}{2} - \frac{y}{3}\]

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

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

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

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

d) \({x^6} + {x^4} + {x^2}{y^2} + {y^4} - {y^6}\)

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

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

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

\( = \left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right) + \left( {{x^2} + {y^2} - xy} \right)\left( {{x^2} + {y^2} + xy} \right)\)

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

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

CÂU HỎI HOT CÙNG CHỦ ĐỀ