Bài 1:
\(a,=2x^3+3x\\ b,=15x-3x^2-5+x=-3x^2+16x-5\\ c,=x^2+4x+4-3x=x^2+x+4\\ d,=\dfrac{x-1+5}{x+3}=\dfrac{x+4}{x+3}\left(x\ne3\right)\)
Bài 2:
\(a,=x\left(4x+3y\right)\\ b,=\left(x-3\right)\left(x+3\right)\\ c,=\left(x-y\right)\left(x+y\right)+3\left(x+y\right)=\left(x+y\right)\left(x-y+3\right)\\ d,=x\left(x^2+16x+64\right)=x\left(x+8\right)^2\)
Bài 3:
\(a,\Leftrightarrow2x^2-3x-2x^2+x=2\\ \Leftrightarrow-2x=2\Leftrightarrow x=-1\\ b,\Leftrightarrow\left(x+3-3x\right)\left(x+3+3x\right)=0\\ \Leftrightarrow\left(3-2x\right)\left(4x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
