1)${{\left( {{x}^{2}}-4x \right)}^{2}}-2\left( {{x}^{2}}-4x \right)-15=0$
Đặt $t={{x}^{2}}-4x$
Phương trình trở thành:
$\,\,\,\,\,\,\,{{t}^{2}}-2t-15=0$
$\Leftrightarrow \left( t-5 \right)\left( t+3 \right)=0$
$\Leftrightarrow\left[\begin{array}{l}t=5\\t=-3\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{l}x^2-4x=5\\x^2-4x=-3\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{l}x^2-4x-5=0\\x^2-4x+3=0\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{l}\left(x-5\right)\left(x+1\right)=0\\\left(x-3\right)\left(x-1\right)=0\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{l}x=5\\x=-1\\x=3\\x=1\end{array}\right.$
Vậy tập nghiệm $S=\left\{ 5\,;\,3\,;\,1\,;\,-1 \right\}$
2)${{\left( 2{{x}^{2}}+3x-1 \right)}^{2}}-5\left( 2{{x}^{2}}+3x+3 \right)+24=0$
Đặt $t=2{{x}^{2}}+3x-1$
Phương trình trở thành:
$\,\,\,\,\,\,\,{{t}^{2}}-5\left( t+4 \right)+24=0$
$\Leftrightarrow {{t}^{2}}-5t+4=0$
$\Leftrightarrow \left( t-4 \right)\left( t-1 \right)=0$
$\Leftrightarrow\left[\begin{array}{l}t=4\\t=1\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{l}2x^2+3x-1=4\\2x^2+3x-1=1\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{l}2x^2+3x-5=0\\2x^2+3x-2=0\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{l}\left(2x+5\right)\left(x-1\right)=0\\\left(2x-1\right)\left(x+2\right)=0\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{l}x=-\dfrac{5}{2}\\x=1\\x=\dfrac{1}{2}\\x=-2\end{array}\right.$
Vậy tập nghiệm $S=\left\{ 1\,;\,\dfrac{1}{2}\,;\,-2\,;\,-\dfrac{5}{2} \right\}$
3) $\left( x+1 \right)\left( x+2 \right)\left( x+4 \right)\left( x+5 \right)=40$
$\Leftrightarrow \left( x+1 \right)\left( x+5 \right)\left( x+2 \right)\left( x+4 \right)=40$
$\Leftrightarrow \left( {{x}^{2}}+6x+5 \right)\left( {{x}^{2}}+6x+8 \right)=40$
Đặt $t={{x}^{2}}+6x+5$
Phương trình trở thành:
$\,\,\,\,\,\,\,t\left( t+3 \right)=40$
$\Leftrightarrow {{t}^{2}}+3t-40=0$
$\Leftrightarrow \left( t-5 \right)\left( t+8 \right)=0$
$\Leftrightarrow\left[\begin{array}{l}t=5\\t=-8\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{l}x^2+6x+5=5\\x^2+6x+5=-8\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{l}x^2+6x=0\\x^2+6x+13=0\,\,\,\,\,\left(\text{ VN }\right)\end{array}\right.$
$\Leftrightarrow x\left(x+6\right)=0$
$\Leftrightarrow\left[\begin{array}{l}x=0\\x=-6\end{array}\right.$
Vậy tập nghiệm $S=\left\{ 0\,;\,-6 \right\}$
4) $\left( x+1 \right)\left( x+2 \right)\left( x+3 \right)\left( x+4 \right)=24$
$\Leftrightarrow \left( x+1 \right)\left( x+4 \right)\left( x+2 \right)\left( x+3 \right)=24$
$\Leftrightarrow \left( {{x}^{2}}+5x+4 \right)\left( {{x}^{2}}+5x+6 \right)=24$
Đặt $t={{x}^{2}}+5x+4$
Phương trình trở thành:
$\,\,\,\,\,\,\,t\left( t+2 \right)=24$
$\Leftrightarrow {{t}^{2}}+2t-24=0$
$\Leftrightarrow \left( t-4 \right)\left( t+6 \right)=0$
$\Leftrightarrow\left[\begin{array}{l}t=4\\t=-6\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{l}x^2+5x+4=4\\x^2+5x+4=-6\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{l}x^2+5x=0\\x^2+5x+10=0\,\,\,\,\,\left(\text{ VN }\right)\end{array}\right.$
$\Leftrightarrow x\left(x+5\right)=0$
$\Leftrightarrow\left[\begin{array}{l}x=0\\x=-5\end{array}\right.$
Vậy tập nghiệm $S=\left\{ 0\,;\,-5 \right\}$