a: x(x+1)(x-1)(x+2)=24
=>\(\left(x^2+x\right)\left(x^2+x-2\right)=24\)
=>\(\left(x^2+x\right)^2-2\left(x^2+x\right)-24=0\)
=>\(\left(x^2+x-6\right)\left(x^2+x+4\right)=0\)
mà \(x^2+x+4=\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}>=\dfrac{15}{4}>0\forall x\)
nên \(x^2+x-6=0\)
=>(x+3)(x-2)=0
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
b: \(\left|x-2\right|\left(x-1\right)\left(x+1\right)\left(x+2\right)=4\)(1)
TH1: x<2
(1) sẽ trở thành:
\(-\left(x-2\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)=4\)
=>\(\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)=-4\)
=>\(\left(x^2-4\right)\left(x^2-1\right)+4=0\)
=>\(x^4-5x^2+8=0\)
=>\(x^4-2\cdot x^2\cdot\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{7}{4}=0\)
=>\(\left(x^2-\dfrac{5}{2}\right)^2+\dfrac{7}{4}=0\)(vô lý)
TH2: x>=2
(1) sẽ trở thành:
\(\left(x-2\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)=4\)
=>\(\left(x^2-1\right)\left(x^2-4\right)=4\)
=>\(x^4-5x^2+4-4=0\)
=>\(x^4-5x^2=0\)
=>\(x^2\left(x^2-5\right)=0\)
=>\(\left[{}\begin{matrix}x^2=0\\x^2-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x^2=5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\sqrt{5}\left(nhận\right)\\x=-\sqrt{5}\left(loại\right)\end{matrix}\right.\)
c: xy-2x+3y=21
=>\(xy-2x+3y-6=15\)
=>\(x\left(y-2\right)+3\left(y-2\right)=15\)
=>\(\left(x+3\right)\left(y-2\right)=15\)
mà x+3>3(Do x là số nguyên dương)
nên \(\left(x+3;y-2\right)\in\left\{\left(5;3\right);\left(15;1\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(2;5\right);\left(12;3\right)\right\}\)
