Giải thích các bước giải:
a,
Ta có:
\(\begin{array}{l}
\left( {{x^2} + 9} \right)\left( {{x^2} + 9x} \right) = 22{\left( {x - 1} \right)^2}\\
\Leftrightarrow {x^4} + 9{x^3} + 9{x^2} + 81x = 22\left( {{x^2} - 2x + 1} \right)\\
\Leftrightarrow {x^4} + 9{x^3} - 13{x^2} + 125x - 22 = 0\\
\Leftrightarrow \left( {{x^4} + 11{x^3} - 2{x^2}} \right) - \left( {2{x^3} + 22{x^2} - 4x} \right) + \left( {11{x^2} + 121x - 22} \right) = 0\\
\Leftrightarrow {x^2}\left( {{x^2} + 11x - 2} \right) - 2x\left( {{x^2} + 11x - 2} \right) + 11\left( {{x^2} + 11x - 2} \right) = 0\\
\Leftrightarrow \left( {{x^2} - 2x + 11} \right)\left( {{x^2} + 11x - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} - 2x + 11 = 0\,\,\,\,\,\left( {vn} \right)\\
{x^2} + 11x - 2 = 0
\end{array} \right.\\
\Leftrightarrow {x^2} + 11x - 2 = 0\\
\Leftrightarrow x = \frac{{ - 11 \pm \sqrt {129} }}{2}
\end{array}\)
b,
ĐKXĐ: \(\left\{ \begin{array}{l}
{y^2} - 3x \ge 0\\
{x^2} + 8y \ge 0
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
\sqrt {{y^2} - 3x} + \sqrt {{x^2} + 8y} = 5\\
x\left( {x - 3} \right) + y\left( {y + 8} \right) = 13
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt {{y^2} - 3x} = 5 - \sqrt {{x^2} + 8y} \\
\left( {{x^2} + 8y} \right) + \left( {{y^2} - 3x} \right) = 13
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt {{y^2} - 3x} = 5 - \sqrt {{x^2} + 8y} \\
\left( {{x^2} + 8y} \right) + {\left( {5 - \sqrt {{x^2} + 8y} } \right)^2} = 13
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt {{y^2} - 3x} = 5 - \sqrt {{x^2} + 8y} \\
\left( {{x^2} + 8y} \right) + \left( {{x^2} + 8y} \right) - 10\sqrt {{x^2} + 8y} + 25 = 13
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\sqrt {{y^2} - 3x} = 5 - \sqrt {{x^2} + 8y} \\
\left( {{x^2} + 8y} \right) - 5\sqrt {{x^2} + 8y} + 6 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\sqrt {{x^2} + 8y} = 2\\
\sqrt {{y^2} - 3x} = 3
\end{array} \right.\\
\left\{ \begin{array}{l}
\sqrt {{x^2} + 8y} = 3\\
\sqrt {{y^2} - 3x} = 2
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{x^2} + 8y = 4\\
{y^2} - 3x = 9
\end{array} \right.\\
\left\{ \begin{array}{l}
{x^2} + 8y = 9\\
{y^2} - 3x = 4
\end{array} \right.
\end{array} \right.\\
TH1:\,\,\left\{ \begin{array}{l}
{x^2} + 8y = 4\,\,\,\left( 1 \right)\\
{y^2} - 3x = 9\,\,\,\left( 2 \right)
\end{array} \right.\\
\left( 2 \right) \Leftrightarrow x = \frac{{{y^2} - 9}}{3}\\
\left( 1 \right) \Leftrightarrow {\left( {\frac{{{y^2} - 9}}{3}} \right)^2} + 8y = 4\\
\Leftrightarrow \frac{{{y^4}}}{9} - 2{y^2} + 9 + 8y = 4\\
\Leftrightarrow {y^4} - 18{y^2} + 72y + 35 = 0
\end{array}\)
