1. Cho a^2+b^2+c^2+3=2.(a+b+c). Chứng minh : a=b=a=1
2.Cho (a+b+c)^2=3.(ab+ac+bc).Chứng minh a=b=c
3.Cho (a-b)^2 + (b-c)^2 + (c-a)^2 = (a+b-2c)^2 + (b+c-2a)^2 + (c+a-2b)^2. Chứng minh a=b=c
4.Cho a,b,c,d là các số khác 0 và
(a+b+c+d). (a-b-c+d) = (a-b+c-d).(a+b-c-d). Chứng minh a/c = b/d
Giải thích các bước giải:
\(\begin{array}{l}
1,\\
{a^2} + {b^2} + {c^2} + 3 = 2\left( {a + b + c} \right)\\
\Leftrightarrow {a^2} + {b^2} + {c^2} + 3 – 2\left( {a + b + c} \right) = 0\\
\Leftrightarrow \left( {{a^2} – 2a + 1} \right) + \left( {{b^2} – 2b + 1} \right) + \left( {{c^2} – 2c + 1} \right) = 0\\
\Leftrightarrow {\left( {a – 1} \right)^2} + {\left( {b – 1} \right)^2} + {\left( {c – 1} \right)^2} = 0\\
\left. \begin{array}{l}
{\left( {a – 1} \right)^2} \ge 0,\,\,\,\forall a\\
{\left( {b – 1} \right)^2} \ge 0,\,\,\,\forall b\\
{\left( {c – 1} \right)^2} \ge 0,\,\,\,\forall c
\end{array} \right\} \Rightarrow {\left( {a – 1} \right)^2} + {\left( {b – 1} \right)^2} + {\left( {c – 1} \right)^2} \ge 0,\,\,\,\forall a,b,c\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {a – 1} \right)^2} = 0\\
{\left( {b – 1} \right)^2} = 0\\
{\left( {c – 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = 1\\
b = 1\\
c = 1
\end{array} \right. \Rightarrow a = b = c = 1\\
2,\\
{\left( {a + b + c} \right)^2} = 3\left( {ab + bc + ca} \right)\\
\Leftrightarrow {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca = 3\left( {ab + bc + ca} \right)\\
\Leftrightarrow {a^2} + {b^2} + {c^2} – ab – bc – ca = 0\\
\Leftrightarrow 2{a^2} + 2{b^2} + 2{c^2} – 2ab – 2bc – 2ca = 0\\
\Leftrightarrow \left( {{a^2} – 2ab + {b^2}} \right) + \left( {{b^2} – 2bc + {c^2}} \right) + \left( {{c^2} – 2ca + {a^2}} \right) = 0\\
\Leftrightarrow {\left( {a – b} \right)^2} + {\left( {b – c} \right)^2} + {\left( {c – a} \right)^2} = 0\\
\left. \begin{array}{l}
{\left( {a – b} \right)^2} \ge 0,\,\,\forall a,b\\
{\left( {b – c} \right)^2} \ge 0,\,\,\,\forall b,c\\
{\left( {c – a} \right)^2} \ge 0,\,\,\,\forall c,a
\end{array} \right\} \Rightarrow {\left( {a – b} \right)^2} + {\left( {b – c} \right)^2} + {\left( {c – a} \right)^2} \ge 0,\,\,\forall a,b,c\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {a – b} \right)^2} = 0\\
{\left( {b – c} \right)^2} = 0\\
{\left( {c – a} \right)^2} = 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a = b\\
b = c\\
c = a
\end{array} \right. \Rightarrow a = b = c
\end{array}\)