Phân tích thành nhân tử : 1 , (a+b+c)(ab+bc+ca)-abc 2, 4a(a+b)(a+b+c)(a+c)+b2c2 3, a(b2+c2)+b(c2+a2)+c(a2+b2)+3abc 4, ab(a+b)-bc(b+c)-ca(c-a)

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Phân tích thành nhân tử :
1 , (a+b+c)(ab+bc+ca)-abc
2, 4a(a+b)(a+b+c)(a+c)+b2c2
3, a(b2+c2)+b(c2+a2)+c(a2+b2)+3abc
4, ab(a+b)-bc(b+c)-ca(c-a)

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Mary 2 tháng 2021-08-07T18:57:46+00:00 1 Answers 3 views 0

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    2021-08-07T18:59:28+00:00

    Giải thích các bước giải:

    \(\begin{array}{l}
    1,\\
    \left( {a + b + c} \right)\left( {ab + bc + ca} \right) – abc\\
     = \left( {{a^2}b + abc + {a^2}c + a{b^2} + {b^2}c + abc + abc + b{c^2} + {c^2}a} \right) – abc\\
     = {a^2}b + a{b^2} + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc\\
     = \left( {{a^2}b + a{b^2}} \right) + \left( {{a^2}c + 2abc + {b^2}c} \right) + \left( {a{c^2} + b{c^2}} \right)\\
     = ab\left( {a + b} \right) + c.\left( {{a^2} + 2ab + {b^2}} \right) + {c^2}\left( {a + b} \right)\\
     = ab\left( {a + b} \right) + c.{\left( {a + b} \right)^2} + {c^2}\left( {a + b} \right)\\
     = \left( {a + b} \right).\left( {ab + c.\left( {a + b} \right) + {c^2}} \right)\\
     = \left( {a + b} \right).\left( {ab + ca + cb + {c^2}} \right)\\
     = \left( {a + b} \right).\left( {a\left( {b + c} \right) + c\left( {b + c} \right)} \right)\\
     = \left( {a + b} \right).\left( {b + c} \right).\left( {c + a} \right)\\
    2,\\
    4a\left( {a + b} \right)\left( {a + b + c} \right).\left( {a + c} \right) + {b^2}{c^2}\\
     = 4.\left[ {a.\left( {a + b + c} \right)} \right].\left[ {\left( {a + b} \right)\left( {a + c} \right)} \right] + {b^2}{c^2}\\
     = 4.\left( {{a^2} + ab + ac} \right).\left( {{a^2} + ab + ac + bc} \right) + {b^2}{c^2}\\
     = 4.{\left( {{a^2} + ab + ac} \right)^2} + 4bc.\left( {{a^2} + ab + ac} \right) + {b^2}{c^2}\\
     = {\left[ {2.\left( {{a^2} + ab + ac} \right)} \right]^2} + 2.2.\left( {{a^2} + ab + ac} \right).bc + {\left( {bc} \right)^2}\\
     = {\left( {2{a^2} + 2ab + 2ac + bc} \right)^2}\\
    3,\\
    a\left( {{b^2} + {c^2}} \right) + b\left( {{c^2} + {a^2}} \right) + c\left( {{a^2} + {b^2}} \right) + 3abc\\
     = a{b^2} + a{c^2} + b{c^2} + b{a^2} + c{a^2} + c{b^2} + 3abc\\
     = \left( {a{b^2} + {a^2}b + abc} \right) + \left( {a{c^2} + c{a^2} + abc} \right) + \left( {b{c^2} + c{b^2} + abc} \right)\\
     = ab\left( {a + b + c} \right) + ac\left( {a + b + c} \right) + bc\left( {a + b + c} \right)\\
     = \left( {a + b + c} \right).\left( {ab + ac + bc} \right)\\
    4,\\
    ab\left( {a + b} \right) – bc\left( {b + c} \right) – ca\left( {c – a} \right)\\
     = ab\left( {a + b} \right) – {b^2}c – b{c^2} – {c^2}a + c{a^2}\\
     = ab\left( {a + b} \right) + \left( {c{a^2} – {b^2}c} \right) – \left( {b{c^2} + {c^2}a} \right)\\
     = ab\left( {a + b} \right) + c.\left( {{a^2} – {b^2}} \right) – {c^2}\left( {a + b} \right)\\
     = ab\left( {a + b} \right) + c.\left( {a – b} \right)\left( {a + b} \right) – {c^2}\left( {a + b} \right)\\
     = \left( {a + b} \right).\left( {ab + ca – cb – {c^2}} \right)\\
     = \left( {a + b} \right).\left( {a.\left( {b + c} \right) – c\left( {b + c} \right)} \right)\\
     = \left( {a + b} \right).\left( {b + c} \right)\left( {a – c} \right)
    \end{array}\)

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