đề bài : Chứng minh các hằng đẳng thức sau:
1. x^6 +3x^2y^2 +y^6 =1 với x^2 + y^2=1
2. x^4 +x^2y^2 +y^4 = a^2 – b^2 với x^2 + y^2 =a , xy=b
3. ( A^3 +b^3 – a^3b^3)^3 + 27a^6b^6=0 với ab=a+b
4.p^2 + (p – a)^2 + (p -b)^2 + (p -c)^2=a^2 + b^2 +c^2 với a + b + c = 2p
Đáp án:
$\begin{array}{l}
1)Do:{x^2} + {y^2} = 1\\
\Rightarrow {x^6} + 3{x^2}{y^2} + {y^6}\\
= {x^6} + {y^6} + 3{x^2}{y^2}\\
= {\left( {{x^2}} \right)^3} + {\left( {{y^2}} \right)^3} + 3{x^2}{y^2}\\
= \left( {{x^2} + {y^2}} \right)\left( {{x^4} – {x^2}{y^2} + {y^4}} \right) + 3{x^2}{y^2}\\
= 1.\left( {{x^4} – {x^2}{y^2} + {y^4}} \right) + 3{x^2}{y^2}\\
= {x^4} + 2{x^2}{y^2} + {y^4}\\
= {\left( {{x^2} + {y^2}} \right)^2}\\
= {1^2} = 1\\
Vay\,{x^6} + 3{x^2}{y^2} + {y^6} = 1\\
2)\\
{x^4} + {x^2}{y^2} + {y^4}\\
= {x^4} + 2{x^2}{y^2} + {y^4} – {x^2}{y^2}\\
= {\left( {{x^2} + {y^2}} \right)^2} – {x^2}{y^2}\\
= {\left( {{x^2} + {y^2}} \right)^2} – {\left( {xy} \right)^2}\\
= {a^2} – {b^2}\\
\left( {do:\left\{ \begin{array}{l}
{x^2} + {y^2} = a\\
xy = b
\end{array} \right.} \right)\\
3)\\
{\left( {{a^3} + {b^3} – {a^3}{b^3}} \right)^3} + 27{a^6}{b^6}\\
= {\left( {{a^3} + {b^3} – {{\left( {a + b} \right)}^3}} \right)^3} + 27{a^6}{b^6}\\
= {\left( {{a^3} + {b^3} – {a^3} – 3{a^2}b – 3a{b^2} – {b^3}} \right)^3} + 27{a^6}{b^6}\\
= {\left( { – 3{a^2}b – 3a{b^2}} \right)^3} + 27{a^6}{b^6}\\
= {\left[ { – 3ab\left( {a + b} \right)} \right]^3} + 27{a^6}{b^6}\\
= – 27{a^3}{b^3}.{\left( {a + b} \right)^3} + 27{a^6}{b^6}\\
= – 27{a^3}{b^3}.{\left( {ab} \right)^3} + 27{a^6}{b^6}\\
= – 27{a^6}{b^6} + 27{a^6}{b^6}\\
= 0\\
4)\\
a + b + c = 2p\\
\Rightarrow {p^2} + {\left( {p – a} \right)^2} + {\left( {p – b} \right)^2} + {\left( {p – c} \right)^2}\\
= \dfrac{1}{4}{\left( {a + b + c} \right)^2} + {\left( {\dfrac{1}{2}b + \dfrac{1}{2}c – \dfrac{1}{2}a} \right)^2}\\
+ {\left( {\dfrac{1}{2}a + \dfrac{1}{2}c – \dfrac{1}{2}b} \right)^2} + {\left( {\dfrac{1}{2}a + \dfrac{1}{2}b – \dfrac{1}{2}c} \right)^2}\\
= \dfrac{1}{4}\left( {{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ac} \right)\\
+ \dfrac{1}{4}\left( {{a^2} + {b^2} + {c^2} + 2bc – 2ac – 2ab} \right)\\
+ \dfrac{1}{4}\left( {{a^2} + {b^2} + {c^2} – 2ab – 2bc + 2ac} \right)\\
+ \dfrac{1}{4}\left( {{a^2} + {b^2} + {c^2} + 2ab – 2bc – 2ac} \right)\\
= {a^2} + {b^2} + {c^2}\\
Vay\,{p^2} + {\left( {p – a} \right)^2} + {\left( {p – b} \right)^2} + {\left( {p – c} \right)^2}\\
= {a^2} + {b^2} + {c^2}
\end{array}$