Question

Đã trả lời bởi chuyên gia

Answer 1

\(\begin{array}{l}
a){x^3} - 25 - (x + 5) = 0\\
\Leftrightarrow {x^3} - 25 - x - 5 = 0\\
\Leftrightarrow {x^3} - x - 30 = 0\\
\Leftrightarrow x = 3,21\\
vay:x = 3,21\\
b){x^3} - 8 - (x - 2)(x - 12) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 2{\rm{x}} + 4} \right) - \left( {x - 2} \right)\left( {x - 12} \right) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 2{\rm{x}} + 4 - x + 12} \right) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + x + 16} \right) = 0\\
\Leftrightarrow x - 2 = 0\left( {do:{x^2} + x + 16 > 0} \right)\\
\Leftrightarrow x = 2\\
Vay:x = 2\\
c){(3x - 1)^2} - {(x + 5)^2} = 0\\
\Leftrightarrow \left( {3{\rm{x - 1 - x - 5}}} \right)\left( {3{\rm{x}} - 1 + x + 5} \right) = 0\\
\Leftrightarrow \left( {2{\rm{x}} - 6} \right)\left( {4{\rm{x}} + 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2{\rm{x}} - 6 = 0\\
4{\rm{x}} + 4 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = - 1
\end{array} \right.\\
Vay:S = \{ 3; - 1\} \\
d){x^2} - 5x - 14 = 0\\
\Leftrightarrow {x^2} - 7{\rm{x}} + 2{\rm{x}} - 14 = 0\\
\Leftrightarrow x\left( {x - 7} \right) + 2\left( {x - 7} \right) = 0\\
\Leftrightarrow \left( {x - 7} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 7 = 0\\
x + 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 7\\
x = - 2
\end{array} \right.\\
vay:S = \{ 7; - 2\}
\end{array}\)

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Answer 2

\(\begin{array}{l}
a){x^3} - 25 - (x + 5) = 0\\
\Leftrightarrow {x^3} - 25 - x - 5 = 0\\
\Leftrightarrow {x^3} - x - 30 = 0\\
\Leftrightarrow x = 3,21\\
vay:x = 3,21\\
b){x^3} - 8 - (x - 2)(x - 12) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 2{\rm{x}} + 4} \right) - \left( {x - 2} \right)\left( {x - 12} \right) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 2{\rm{x}} + 4 - x + 12} \right) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + x + 16} \right) = 0\\
\Leftrightarrow x - 2 = 0\left( {do:{x^2} + x + 16 > 0} \right)\\
\Leftrightarrow x = 2\\
Vay:x = 2\\
c){(3x - 1)^2} - {(x + 5)^2} = 0\\
\Leftrightarrow \left( {3{\rm{x - 1 - x - 5}}} \right)\left( {3{\rm{x}} - 1 + x + 5} \right) = 0\\
\Leftrightarrow \left( {2{\rm{x}} - 6} \right)\left( {4{\rm{x}} + 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2{\rm{x}} - 6 = 0\\
4{\rm{x}} + 4 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = - 1
\end{array} \right.\\
Vay:S = \{ 3; - 1\} \\
d){x^2} - 5x - 14 = 0\\
\Leftrightarrow {x^2} - 7{\rm{x}} + 2{\rm{x}} - 14 = 0\\
\Leftrightarrow x\left( {x - 7} \right) + 2\left( {x - 7} \right) = 0\\
\Leftrightarrow \left( {x - 7} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 7 = 0\\
x + 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 7\\
x = - 2
\end{array} \right.\\
vay:S = \{ 7; - 2\}
\end{array}\)

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