Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
{\left( {\frac{{{{\rm{x}}^2} + 4{\rm{x}}}}{{{\rm{x}} + 1}}} \right)^2} + 3x = 19 + \frac{9}{{x + 1}}\\
dk:x \ne - 1\\
\Leftrightarrow \frac{{{{\left( {{x^2} + 4x} \right)}^2}}}{{{{\left( {x + 1} \right)}^2}}} + \frac{{3x{{\left( {x + 1} \right)}^2}}}{{{{\left( {x + 1} \right)}^2}}} = \frac{{19{{\left( {x + 1} \right)}^2}}}{{{{\left( {x + 1} \right)}^2}}} + \frac{{9\left( {x + 1} \right)}}{{{{\left( {x + 1} \right)}^2}}}\\
\Leftrightarrow {\left( {{x^2} + 4x} \right)^2} + 3x{\left( {x + 1} \right)^2} = 19{\left( {x + 1} \right)^2} + 9\left( {x + 1} \right)\\
\Leftrightarrow {x^4} + 8{x^3} + 16{x^2} + 3x\left( {{x^2} + 2x + 1} \right) = 19\left( {{x^2} + 2x + 1} \right) + 9x + 9\\
\Leftrightarrow {x^4} + 8{x^3} + 16{x^2} + 3{x^3} + 6{x^2} + 3x - 19{x^2} - 38x - 19 - 9x - 9 = 0\\
\Leftrightarrow {x^4} + 11{x^3} + 3{x^2} - 44x - 28 = 0\\
\Leftrightarrow {x^4} - 2{x^3} + 13{x^3} - 26{x^2} + 29{x^2} - 58x + 14x - 28 = 0\\
\Leftrightarrow {x^3}\left( {x - 2} \right) + 13{x^2}\left( {x - 2} \right) + 29x\left( {x - 2} \right) + 14\left( {x - 2} \right) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {{x^3} + 13{x^2} + 29x + 14} \right) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left[ {{x^3} + 2{x^2} + 11{x^2} + 22x + 7x + 14} \right] = 0\\
\Leftrightarrow \left( {x - 2} \right)\left[ {{x^2}\left( {x + 2} \right) + 11x\left( {x + 2} \right) + 7\left( {x + 2} \right)} \right] = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {x + 2} \right)\left( {{x^2} + 11x + 7} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x + 2 = 0\\
{x^2} + 11x + 7 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = - 2\\
x = \frac{{ - 11 + \sqrt {93} }}{2}\\
x = \frac{{ - 11 - \sqrt {93} }}{2}
\end{array} \right.\left( {tm} \right)\\
vay:S = \{ 2; - 2;\frac{{ - 11 + \sqrt {93} }}{2};\frac{{ - 11 - \sqrt {93} }}{2}\}
\end{array}\]

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Answer 2

\[\begin{array}{l}
{\left( {\frac{{{{\rm{x}}^2} + 4{\rm{x}}}}{{{\rm{x}} + 1}}} \right)^2} + 3x = 19 + \frac{9}{{x + 1}}\\
dk:x \ne - 1\\
\Leftrightarrow \frac{{{{\left( {{x^2} + 4x} \right)}^2}}}{{{{\left( {x + 1} \right)}^2}}} + \frac{{3x{{\left( {x + 1} \right)}^2}}}{{{{\left( {x + 1} \right)}^2}}} = \frac{{19{{\left( {x + 1} \right)}^2}}}{{{{\left( {x + 1} \right)}^2}}} + \frac{{9\left( {x + 1} \right)}}{{{{\left( {x + 1} \right)}^2}}}\\
\Leftrightarrow {\left( {{x^2} + 4x} \right)^2} + 3x{\left( {x + 1} \right)^2} = 19{\left( {x + 1} \right)^2} + 9\left( {x + 1} \right)\\
\Leftrightarrow {x^4} + 8{x^3} + 16{x^2} + 3x\left( {{x^2} + 2x + 1} \right) = 19\left( {{x^2} + 2x + 1} \right) + 9x + 9\\
\Leftrightarrow {x^4} + 8{x^3} + 16{x^2} + 3{x^3} + 6{x^2} + 3x - 19{x^2} - 38x - 19 - 9x - 9 = 0\\
\Leftrightarrow {x^4} + 11{x^3} + 3{x^2} - 44x - 28 = 0\\
\Leftrightarrow {x^4} - 2{x^3} + 13{x^3} - 26{x^2} + 29{x^2} - 58x + 14x - 28 = 0\\
\Leftrightarrow {x^3}\left( {x - 2} \right) + 13{x^2}\left( {x - 2} \right) + 29x\left( {x - 2} \right) + 14\left( {x - 2} \right) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {{x^3} + 13{x^2} + 29x + 14} \right) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left[ {{x^3} + 2{x^2} + 11{x^2} + 22x + 7x + 14} \right] = 0\\
\Leftrightarrow \left( {x - 2} \right)\left[ {{x^2}\left( {x + 2} \right) + 11x\left( {x + 2} \right) + 7\left( {x + 2} \right)} \right] = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {x + 2} \right)\left( {{x^2} + 11x + 7} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x + 2 = 0\\
{x^2} + 11x + 7 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = - 2\\
x = \frac{{ - 11 + \sqrt {93} }}{2}\\
x = \frac{{ - 11 - \sqrt {93} }}{2}
\end{array} \right.\left( {tm} \right)\\
vay:S = \{ 2; - 2;\frac{{ - 11 + \sqrt {93} }}{2};\frac{{ - 11 - \sqrt {93} }}{2}\}
\end{array}\]

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