Toán Lớp 9: giải bằng cách đặt ẩn phụ
(x^2+4x+8)^2+3x(x^2+4x+8)+2x^2=0
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Toán Lớp 9: giải bằng cách đặt ẩn phụ
(x^2+4x+8)^2+3x(x^2+4x+8)+2x^2=0
Comments ( 1 )
Đặt:{x^2} + 4x + 8 = a\\
\Leftrightarrow {a^2} + 3.x.a + 2.{x^2} = 0\\
\Leftrightarrow {a^2} + 2.x.a + x.a + 2{x^2} = 0\\
\Leftrightarrow a.\left( {a + 2x} \right) + x.\left( {a + 2x} \right) = 0\\
\Leftrightarrow \left( {a + 2x} \right)\left( {a + x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} + 4x + 8 + 2x = 0\\
{x^2} + 4x + 8 + x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} + 6x + 8 = 0\\
{x^2} + 5x + 8 = 0\left( {vn} \right)
\end{array} \right.\\
\Leftrightarrow \left[ {\left( {x + 2} \right)\left( {x + 4} \right) = 0} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = – 2\\
x = – 4
\end{array} \right.\\
Vậy\,x = – 2;x = – 4
\end{array}$