Bài 1:
\(a,=\left(8x^2+6x\right)-\left(20x+15\right)=0\\ \Leftrightarrow2x\left(4x+3\right)-5\left(4x+3\right)=0\\ \Leftrightarrow\left(2x-5\right)\left(4x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
Câu 2:
\(a,\Leftrightarrow\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+2\right)\left(x+4\right)\right]=40\\ \Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)
Đặt \(x^2+6x+5=t\)
\(\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\\ \Leftrightarrow t\left(t+3\right)=40\\ \Leftrightarrow t^2+3t-40=0\\ \Leftrightarrow\left(t^2+8t\right)-\left(5t+40\right)=0\\ \Leftrightarrow t\left(t+8\right)-5\left(t+8\right)=0\\ \Leftrightarrow\left(t-5\right)\left(t+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=5\\t=8\end{matrix}\right.\)
Với
\(t=5\Leftrightarrow x^2+6x+5=5\\ \Leftrightarrow x\left(x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
\(t=-8\Leftrightarrow x^2+6x+5=-8\\ \Leftrightarrow x^2+6x+13=0\\ \Leftrightarrow\left(x^2+6x+9\right)+4=0\\ \Leftrightarrow\left(x+3\right)^2+4=0\left(vôlí\right)\)
\(1,\\ a,=8x^2-20x+6x-15=\left(2x-5\right)\left(4x+3\right)\\ b,\Leftrightarrow\left(2x^3-x^2+2x^2-x+8x-4+3\right)⋮\left(2x-1\right)\\ \Leftrightarrow\left[\left(2x-1\right)\left(x^2+x+4\right)+3\right]⋮\left(2x-1\right)\\ \Leftrightarrow2x-1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{-1;0;1;2\right\}\\ 2,\\ a,\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\\ \Leftrightarrow\left(x^2+6x\right)^2+13\left(x^2+6x\right)+40=40\\ \Leftrightarrow\left(x^2+6x\right)^2+13\left(x^2+6x\right)=0\\ \Leftrightarrow\left(x^2+6x\right)\left(x^2+6x+13\right)=0\\ \Leftrightarrow x\left(x+6\right)\left[\left(x+3\right)^2+4\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\left[\left(x+3\right)^2+4>0\right]\)
\(b,\Leftrightarrow y\left(x-1\right)-x^2+1=3\\ \Leftrightarrow y\left(x-1\right)-\left(x-1\right)\left(x+1\right)=3\\ \Leftrightarrow\left(x-1\right)\left(y-x-1\right)=3=1\cdot3=\left(-1\right)\left(-3\right)\)
\(TH_1:\left\{{}\begin{matrix}x-1=1\\y-x-1=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\\ TH_2:\left\{{}\begin{matrix}x-1=3\\y-x-1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=6\end{matrix}\right.\\ TH_3:\left\{{}\begin{matrix}x-1=-1\\y-x-1=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-2\end{matrix}\right.\\ TH_4:\left\{{}\begin{matrix}x-1=-3\\y-x-1=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-2\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{...\right\}\)
