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Ben Gia
Ben Gia

Ben Gia

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Ben Gia
Asked: Tháng Mười 26, 20202020-10-26T01:38:14+00:00 2020-10-26T01:38:14+00:00In: Môn Toán

Cho a+b+c+d =0 a^3+b^3+c^3+d^3=3(ab-cd)(c+d)

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Cho a+b+c+d =0
a^3+b^3+c^3+d^3=3(ab-cd)(c+d)

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    1. Amity

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      Amity
      2020-10-26T01:39:41+00:00Added an answer on Tháng Mười 26, 2020 at 1:39 sáng

      Giải thích các bước giải:

       Ta có:

      \(\begin{array}{l}
      a + b + c + d = 0 \Rightarrow a + b =  – \left( {c + d} \right)\\
      {a^3} + {b^3} + {c^3} + {d^3}\\
       = \left( {{a^3} + 3{a^2}b + 3a{b^2} + {b^3}} \right) + \left( {{c^3} + 3{c^2}d + 3c{d^2} + {d^3}} \right) – \left( {3{a^2}b + 3a{b^2} + 3{c^2}d + 3c{d^2}} \right)\\
       = {\left( {a + b} \right)^3} + {\left( {c + d} \right)^3} – 3ab\left( {a + b} \right) – 3cd\left( {c + d} \right)\\
       = {\left[ { – \left( {c + d} \right)} \right]^3} + {\left( {c + d} \right)^3} – 3ab.\left[ { – \left( {c + d} \right)} \right] – 3cd\left( {c + d} \right)\\
       =  – {\left( {c + d} \right)^3} + {\left( {c + d} \right)^3} + 3ab\left( {c + d} \right) – 3cd\left( {c + d} \right)\\
       = 3ab\left( {c + d} \right) – 3cd\left( {c + d} \right)\\
       = 3.\left( {c + d} \right)\left( {ab – cd} \right)
      \end{array}\)

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