Hãy luôn nhớ cảm ơn và vote 5*
nếu câu trả lời hữu ích nhé!
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
5,\\
a,\\
{\left( {a - 1} \right)^2} + {\left( {b - 1} \right)^2} + {\left( {c - 1} \right)^2} + {\left( {d - 1} \right)^2} \ge 0,\,\,\,\forall a,b,c,d\\
\Leftrightarrow \left( {{a^2} - 2a + 1} \right) + \left( {{b^2} - 2b + 1} \right) + \left( {{c^2} - 2c + 1} \right) + \left( {{d^2} - 2d + 1} \right) \ge 0\\
\Leftrightarrow \left( {{a^2} + {b^2} + {c^2} + {d^2}} \right) - 2\left( {a + b + c + d} \right) + 4 \ge 0\\
\Leftrightarrow {a^2} + {b^2} + {c^2} + {d^2} + 4 \ge 2\left( {a + b + c + d} \right)\\
b,\\
{\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} \ge 0,\,\,\,\forall a,b,c\\
\Leftrightarrow \left( {{a^2} - 2ab + {b^2}} \right) + \left( {{b^2} - 2bc + {c^2}} \right) + \left( {{c^2} - 2ca + {a^2}} \right) \ge 0,\,\,\,\forall a,b,c\\
\Leftrightarrow 2\left( {{a^2} + {b^2} + {c^2}} \right) - 2\left( {ab + bc + ca} \right) \ge 0\\
\Leftrightarrow 2\left( {{a^2} + {b^2} + {c^2}} \right) \ge 2\left( {ab + bc + ca} \right)\\
\Leftrightarrow 3\left( {{a^2} + {b^2} + {c^2}} \right) \ge {a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right)\\
\Leftrightarrow 3\left( {{a^2} + {b^2} + {c^2}} \right) \ge {\left( {a + b + c} \right)^2}\\
c,\\
{\left( {a - b} \right)^2}.\left( {a + b} \right) \ge 0,\,\,\,\forall a,b > 0\\
\Leftrightarrow \left( {a - b} \right)\left( {a - b} \right)\left( {a + b} \right) \ge 0\\
\Leftrightarrow \left( {a - b} \right)\left( {{a^2} - {b^2}} \right) \ge 0\\
\Leftrightarrow {a^3} - a{b^2} - {a^2}b + {b^3} \ge 0\\
\Leftrightarrow {a^3} + {b^3} \ge a{b^2} + {a^2}b\\
\Leftrightarrow {a^3} + {b^3} \ge ab\left( {a + b} \right)\\
d,\\
{\left( {a - b} \right)^2} + {\left( {a - c} \right)^2} + {b^2} + {c^2} \ge 0,\,\,\,\forall a,b,c\\
\Leftrightarrow \left( {{a^2} - 2ab + {b^2}} \right) + \left( {{a^2} - 2ac + {c^2}} \right) + {b^2} + {c^2} \ge 0\\
\Leftrightarrow 2\left( {{a^2} + {b^2} + {c^2}} \right) - 2\left( {ab + ac} \right) \ge 0\\
\Leftrightarrow {a^2} + {b^2} + {c^2} \ge ab + ac\\
\Leftrightarrow {a^2} + {b^2} + {c^2} \ge a\left( {b + c} \right)\\
3,\\
{a^3} + {b^3} = \left( {{a^3} + 3{a^2}b + 3a{b^2} + {b^3}} \right) - 3{a^2}b - 3a{b^2} = {\left( {a + b} \right)^3} - 3ab\left( {a + b} \right)\\
{\left( {x - 1} \right)^3} + {\left( {x + 2} \right)^3} = {\left( {2x + 1} \right)^3}\\
\Leftrightarrow {\left[ {\left( {x - 1} \right) + \left( {x + 2} \right)} \right]^3} - 3.\left( {x - 1} \right)\left( {x + 2} \right).\left[ {\left( {x - 1} \right) + \left( {x + 2} \right)} \right] = {\left( {2x + 1} \right)^3}\\
\Leftrightarrow {\left( {2x + 1} \right)^3} - 3.\left( {x - 1} \right)\left( {x + 2} \right).\left( {2x + 1} \right) = {\left( {2x + 1} \right)^3}\\
\Leftrightarrow 3\left( {x - 1} \right)\left( {x + 2} \right)\left( {2x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x + 2 = 0\\
2x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - 2\\
x = - \frac{1}{2}
\end{array} \right.
\end{array}\)
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