Toán Lớp 8: Giải phương trình: x^3-1+(x-1)(4x+3)=0
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Toán Lớp 8: Giải phương trình: x^3-1+(x-1)(4x+3)=0
Comments ( 2 )
\quad x^3-1+(x-1)(4x+3) = 0\\
\Leftrightarrow (x-1)(x^2 + x + 1) + (x-1)(4x+3) =0\\
\Leftrightarrow (x-1)(x^2 + 5x + 4) = 0\\
\Leftrightarrow (x-1)[(x^2 + x) + (4x + 4)] = 0\\
\Leftrightarrow (x-1)[x(x+1) + 4(x+1)] = 0\\
\Leftrightarrow (x-1)(x+1)(x+4) =0\\
\Leftrightarrow \left[\begin{array}{l}x – 1=0\\x + 1=0\\x + 4=0\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = 1\\x = -1\\x = -4\end{array}\right.\\
\text{Vậy phương trình có các nghiệm là}\ x = -4;x =-1;x = 1
\end{array}\)
{x^3} – 1 + \left( {x – 1} \right)\left( {4x + 3} \right) = 0\\
\Leftrightarrow (x – 1)({x^2} + x + 1) + (x – 1)(4x + 3) = 0\\
\Leftrightarrow (x – 1)({x^2} + 5x + 4) = 0\\
\Leftrightarrow \left( {x – 1} \right)\left( {{x^2} + 4x + x + 4} \right) = 0\\
\Leftrightarrow (x – 1)\left[ {({x^2} + x) + (4x + 4)} \right] = 0\\
\Leftrightarrow (x – 1)\left[ {x(x + 1) + 4(x + 1)} \right] = 0\\
\Leftrightarrow (x – 1)(x + 1)(x + 4) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x – 1 = 0\\
x + 1 = 0\\
x + 4 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = – 1\\
x = – 4
\end{array} \right.\\
\Rightarrow S \in \{ 1; – 1; – 4\} \\
\end{array}\]