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. (

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}$