## Phân tích thành nhân tử :x^2y^2.(y-x)+y^2z^2.(z-y)-z^2x^2.(z-x)

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Phân tích thành nhân tử :x^2y^2.(y-x)+y^2z^2.(z-y)-z^2x^2.(z-x)

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8 tháng 2020-10-31T07:44:56+00:00 1 Answers 70 views 0

$\begin{array}{l} {x^2}{y^2}.(y – x) + {y^2}{z^2}.(z – y) – {z^2}{x^2}.(z – x)\\ = \left( {{x^2}{y^2}.(y – x) – {z^2}{x^2}.(z – x)} \right) + {y^2}{z^2}.(z – y)\\ = {x^2}\left( {{y^3} – x{y^2} – {z^3} + x{z^2}} \right) + {y^2}{z^2}.(z – y)\\ = {x^2}\left( {\left( {{y^3} – {z^3}} \right) – x\left( {{y^2} – {z^2}} \right)} \right) + {y^2}{z^2}.(z – y)\\ = {x^2}\left( {y – z} \right)\left( {{y^2} + yz + {z^2} – xy – xz} \right) – {y^2}{z^2}\left( {y – z} \right)\\ = \left( {y – z} \right)\left( {{x^2}\left( {{y^2} + yz + {z^2} – xy – xz} \right) – {y^2}{z^2}} \right)\\ = \left( {y – z} \right)\left( {{x^2}{y^2} + {x^2}yz + {x^2}{z^2} – {x^3}y – {x^3}z – {y^2}{z^2}} \right)\\ = \left( {y – z} \right)\left( {\left( {{x^2}{y^2} – {y^2}{z^2}} \right) + \left( {{x^2}yz – {x^3}y} \right) + \left( {{x^2}{z^2} – {x^3}z} \right)} \right)\\ = \left( {y – z} \right)\left( {{y^2}\left( {{x^2} – {z^2}} \right) + {x^2}y\left( {z – x} \right) + {x^2}z\left( {z – x} \right)} \right)\\ = \left( {y – z} \right)\left( {z – x} \right)\left( { – {y^2}x – {y^2}z + {x^2}y + {x^2}z} \right)\\ = \left( {y – z} \right)\left( {z – x} \right)\left( {\left( {{x^2}y – {y^2}x} \right) + \left( {{x^2}z – {y^2}z} \right)} \right)\\ = \left( {y – z} \right)\left( {z – x} \right)\left( {xy\left( {x – y} \right) + z\left( {{x^2} – {y^2}} \right)} \right)\\ = \left( {y – z} \right)\left( {z – x} \right)\left( {x – y} \right)\left( {xy + xz + yz} \right) \end{array}$