Đáp án:
d. \(\left\{ \begin{array}{l}
x = - \dfrac{1}{3}\\
y = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.x\left( {x - 4} \right) - {x^2} - 6x = 10\\
\to {x^2} - 4x - {x^2} - 6x = 10\\
\to - 10x = 10\\
\to x = - 1\\
b.{\left[ {x\left( {x + 2} \right)} \right]^2} - 2x\left( {x + 2} \right) = 3\\
\to {\left[ {x\left( {x + 2} \right)} \right]^2} - 2x\left( {x + 2} \right).1 + 1 = 4\\
\to {\left[ {x\left( {x + 2} \right) - 1} \right]^2} = 4\\
\to \left[ \begin{array}{l}
x\left( {x + 2} \right) - 1 = 2\\
x\left( {x + 2} \right) - 1 = - 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} + 2x - 3 = 0\\
{x^2} + 2x + 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left( {x - 1} \right)\left( {x + 3} \right) = 0\\
{\left( {x + 1} \right)^2} = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = - 3\\
x = - 1
\end{array} \right.\\
c.{x^2} + x + \dfrac{1}{4} - \left( {{x^2} + \dfrac{1}{2}x + 6x + 3} \right) = 8\\
\to {x^2} + x + \dfrac{1}{4} - {x^2} - \dfrac{{13}}{2}x - 3 = 8\\
\to - \dfrac{{11}}{2}x = \dfrac{{43}}{4}\\
\to x = - \dfrac{{43}}{{22}}\\
d.9{x^2} + 2.3x.1 + 1 + 4{y^2} - 2.2y.2 + 4 = 0\\
\to {\left( {3x + 1} \right)^2} + {\left( {2y - 2} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
3x + 1 = 0\\
2y - 2 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = - \dfrac{1}{3}\\
y = 1
\end{array} \right.
\end{array}\)
