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

Giải thích các bước giải:

3,

\(\begin{array}{l}
{\left( {a – b} \right)^2} + {\left( {b – c} \right)^2} + {\left( {c – a} \right)^2} = {\left( {a + b – 2c} \right)^2} + {\left( {b + c – 2a} \right)^2} + {\left( {c + a – 2b} \right)^2}\\
 \Leftrightarrow \left[ {{{\left( {a – b} \right)}^2} – {{\left( {a + b – 2c} \right)}^2}} \right] + \left[ {{{\left( {b – c} \right)}^2} – {{\left( {b + c – 2a} \right)}^2}} \right] + \left[ {{{\left( {c – a} \right)}^2} – {{\left( {c + a – 2b} \right)}^2}} \right] = 0\\
 \Leftrightarrow \left[ {\left( {a – b} \right) – \left( {a + b – 2c} \right)} \right].\left[ {\left( {a – b} \right) + \left( {a + b – 2c} \right)} \right] + \left[ {\left( {b – c} \right) – \left( {b + c – 2a} \right)} \right].\left[ {\left( {b – c} \right) + \left( {b + c – 2a} \right)} \right] + \left[ {\left( {c – a} \right) – \left( {c + a – 2b} \right)} \right].\left[ {\left( {c – a} \right) + \left( {c + a – 2b} \right)} \right] = 0\\
 \Leftrightarrow \left( {2c – 2b} \right)\left( {2a – 2c} \right) + \left( {2a – 2c} \right).\left( {2b – 2a} \right) + \left( {2b – 2a} \right).\left( {2c – 2b} \right) = 0\\
 \Leftrightarrow \left( {c – b} \right)\left( {a – c} \right) + \left( {a – c} \right)\left( {b – a} \right) + \left( {b – a} \right).\left( {c – b} \right) = 0\\
 \Leftrightarrow \left( {a – c} \right).\left[ {\left( {c – b} \right) + \left( {b – a} \right)} \right] + \left( {b – a} \right)\left( {c – b} \right) = 0\\
 \Leftrightarrow \left( {a – c} \right).\left( {c – a} \right) + \left( {b – a} \right)\left( {c – b} \right) = 0\\
 \Leftrightarrow  – {a^2} – {c^2} + 2ac + bc – {b^2} – ac + ab = 0\\
 \Leftrightarrow  – {a^2} – {c^2} + ac + bc – {b^2} + ab = 0\\
 \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\\
 \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. \Leftrightarrow a = b = c\\
4,\\
\left( {a + b + c + d} \right)\left( {a – b – c + d} \right) = \left( {a – b + c – d} \right).\left( {a + b – c – d} \right)\\
 \Leftrightarrow \left[ {\left( {a + d} \right) + \left( {b + c} \right)} \right].\left[ {\left( {a + d} \right) – \left( {b + c} \right)} \right] = \left[ {\left( {a – d} \right) – \left( {b – c} \right)} \right].\left[ {\left( {a – d} \right) + \left( {b – c} \right)} \right]\\
 \Leftrightarrow {\left( {a + d} \right)^2} – {\left( {b + c} \right)^2} = {\left( {a – d} \right)^2} – {\left( {b – c} \right)^2}\\
 \Leftrightarrow \left( {{a^2} + 2ad + {d^2}} \right) – \left( {{b^2} + 2bc + {c^2}} \right) = \left( {{a^2} – 2ad + {d^2}} \right) – \left( {{b^2} – 2bc + {c^2}} \right)\\
 \Leftrightarrow 2ad – 2bc =  – 2ad + 2bc\\
 \Leftrightarrow 4ad = 4bc\\
 \Leftrightarrow ad = bc\\
 \Leftrightarrow \frac{a}{c} = \frac{b}{d}
\end{array}\)