Đáp án:
$\begin{array}{l}11)\, x = 1\\12) \, x = -1\\13) \, \left[\begin{array}{l}x =1\\x =-1\end{array}\right.\\14) \, \left[\begin{array}{l}x=\dfrac{1}{2}\\x=-1\end{array}\right.\\15) \, \left[\begin{array}{l}x =-3\\x = \dfrac{2}{5}\end{array}\right.\\16) \, \left[\begin{array}{l}x =-\dfrac{1}{5}\\x =3\end{array}\right.\end{array}$
Giải thích các bước giải:
$\begin{array}{l} 11)\,2x^3(x + 1) - x^2(x+2) = 1\\ \Leftrightarrow x^2[2(x + 1) - (x + 2)] =1\\ \Leftrightarrow x^2(2x + 2 -x - 2) = 1\\ \Leftrightarrow x^3 = 1\\ \Leftrightarrow x = 1\\\\ 12)\,(x+4)(2x - 3) - x(x+3) + 13 = 0\\ \Leftrightarrow 2x^2 - 3x + 8x - 12 - (x^2 + 3x) + 13 = 0\\ \Leftrightarrow x^2 + 2x + 1 = 0\\ \Leftrightarrow (x + 1)^2 = 0\\ \Leftrightarrow x = -1\\\\ 13) \, x^2 - 2x + 1 + (x -1)(x + 3) = 0\\ \Leftrightarrow (x -1)^2 + (x-1)(x+3) = 0\\ \Leftrightarrow (x-1)(x - 1 + x + 3) = 0\\ \Leftrightarrow (x-1)(2x + 2) = 0\\ \Leftrightarrow \left[\begin{array}{l}x - 1 =0\\x + 1=0\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x =1\\x =-1\end{array}\right.\\\\ 14)\, (3x +5)(2x - 1) + 4x^2 - 4x = -1\\ \Leftrightarrow (3x + 5)(2x - 1) + (4x^2 - 4x + 1) = 0\\ \Leftrightarrow (2x + 5)(2x - 1) + (2x -1)^2 = 0\\ \Leftrightarrow (2x - 1)(2x + 5 + 2x -1) = 0\\ \Leftrightarrow (2x - 1)(4x + 4) = 0\\ \Leftrightarrow \left[\begin{array}{l}2x -1 =0\\x + 1=0\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x=\dfrac{1}{2}\\x=-1\end{array}\right.\\\\ 15)\, 2x(x + 3) + (x + 3)(3x - 2) = 0\\ \Leftrightarrow (x+3)(2x + 3x - 2) = 0\\ \Leftrightarrow (x+3)(5x - 2) = 0\\ \Leftrightarrow \left[\begin{array}{l}x+3 =0\\5x -2=0\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x =-3\\x = \dfrac{2}{5}\end{array}\right.\\\\ 16) \, (5x+1)(2x - 7) + (5x + 1)(3x - 8) = 0\\ \Leftrightarrow (5x+1)(2x - 7 + 3x - 8) = 0\\ \Leftrightarrow (5x + 1)(5x - 15) = 0\\ \Leftrightarrow \left[\begin{array}{l}5x + 1 =0\\x-3=0\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x =-\dfrac{1}{5}\\x =3\end{array}\right. \end{array}$
