Đáp án:
8) \(\left[ \begin{array}{l}
x = 5\\
x = - 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)3\left( {{x^2} + 4x + 4} \right) + 4{x^2} - 4x + 1 - 7\left( {{x^2} - 9} \right) = 36\\
\to 3{x^2} + 12x + 12 + 4{x^2} - 4x + 1 - 7{x^2} + 63 = 36\\
\to 8x = - 40\\
\to x = - 5\\
3){x^3} - 3{x^2} + 3x - 1 - {x^3} - 27 + 3{x^2} - 12 = 2\\
\to 3x = 42\\
\to x = 14\\
5){\left( {4 - x} \right)^2} = {\left( {3x + 2} \right)^2}\\
\to \left| {4 - x} \right| = \left| {3x + 2} \right|\\
\to \left[ \begin{array}{l}
4 - x = 3x + 2\\
4 - x = - 3x - 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
4x = 2\\
2x = - 6
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = - 3
\end{array} \right.\\
7)8{x^2} + 2x + 28x + 7 = 0\\
\to 2x\left( {4x + 1} \right) + 7\left( {4x + 1} \right) = 0\\
\to \left( {4x + 1} \right)\left( {2x + 7} \right) = 0\\
\to \left[ \begin{array}{l}
x = - \dfrac{1}{4}\\
x = - \dfrac{7}{2}
\end{array} \right.\\
9)2{x^2} + 3x - 5 = 0\\
\to 2{x^2} - 2x + 5x - 5 = 0\\
\to 2x\left( {x - 1} \right) + 5\left( {x - 1} \right) = 0\\
\to \left( {x - 1} \right)\left( {2x + 5} \right) = 0\\
\to \left[ \begin{array}{l}
x - 1 = 0\\
2x + 5 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = - \dfrac{5}{2}
\end{array} \right.\\
2){x^3} - 1 + x\left( {4 - {x^2}} \right) = 5\\
\to {x^3} - 1 - {x^3} + 4x = 5\\
\to 4x = 6\\
\to x = \dfrac{3}{2}\\
4){x^3} + 27 - x\left( {{x^2} - 4} \right) = 15\\
\to {x^3} + 27 - {x^3} + 4x = 15\\
\to 4x = - 12\\
\to x = - 3\\
6)4{x^2} + 4x + 1 - 4\left( {{x^2} + 4x + 4} \right) = 9\\
\to 4{x^2} + 4x + 1 - 4{x^2} - 16x - 16 = 9\\
\to - 12x = 24\\
\to x = - 2\\
8){x^2} - 2x - 15 = 0\\
\to {x^2} - 5x + 3x - 15 = 0\\
\to x\left( {x - 5} \right) + 3\left( {x - 5} \right) = 0\\
\to \left( {x - 5} \right)\left( {x + 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = 5\\
x = - 3
\end{array} \right.\\
10)\left[ \begin{array}{l}
{x^2} - 4x = 5\\
{x^2} - 4x = 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 5\\
x = - 1\\
x = 2 + \sqrt 7 \\
x = 2 - \sqrt 7
\end{array} \right.
\end{array}\)
