c/m :a) (a^2-b^2)^2+(2ab)^2=(a^2+b^2)^2 b) (a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2

0 bình luận về “c/m :a) (a^2-b^2)^2+(2ab)^2=(a^2+b^2)^2 b) (a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2”

1. Đáp án:

   $\begin{array}{l}
   a){\left( {{a^2} – {b^2}} \right)^2} + {\left( {2ab} \right)^2}\\
    = {a^4} – 2{a^2}{b^2} + {b^4} + 4{a^2}{b^2}\\
    = {a^4} + 2{a^2}{b^2} + {b^4}\\
    = {\left( {{a^2} + {b^2}} \right)^2}\\
   b)\left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right)\\
    = {a^2}{c^2} + {a^2}{d^2} + {b^2}{c^2} + {b^2}{d^2}\\
   {\left( {ac + bd} \right)^2} + {\left( {ad – bc} \right)^2}\\
    = {a^2}{c^2} + 2abcd + {b^2}{d^2} + {a^2}{d^2} – 2abcd + {b^2}{c^2}\\
    = {a^2}{c^2} + {a^2}{d^2} + {b^2}{c^2} + {b^2}{d^2}\\
    \Rightarrow \left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right) = {\left( {ac + bd} \right)^2} + {\left( {ad – bc} \right)^2}
   \end{array}$

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