Đáp án+Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 1:\ \\ a) \ 5x( 4-2x) =20x-10x^{2}\\ b) \ ( 3x-2)( 1+2x) =3x+6x^{2} -2-4x=6x^{2} -x-2\\ c) \ ( 3-x)^{2} -( x+2)( 3-x) =( 3-x)( 3-x-x-2)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =( 3-x)( 1-2x)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3-6x-x+2x^{2} =2x^{2} -7x+3\\ Bài\ 2:\\ ( 2x-3)( -2x+3) +( 1-2x)^{2} =3( x-2)\\ \Leftrightarrow -4x^{2} +6x+6x-9+1-4x+4x^{2} -3x+6=0\\ \Leftrightarrow \ 5x-2=0\\ \Leftrightarrow \ x=\frac{2}{5}\\ Vậy\ \ x=\frac{2}{5}\\ Bài\ 3:\ \\ a) \ AC=\ AB+BC=\ ( 2x+3) +( x-1) =\ 3x+2\\ b) \ Có\ CD=x-1\\ S_{BCEF} =\frac{BC+EF}{2} .CD=\frac{( x-1) +( 2x+3)}{2} .\ ( x-1)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \frac{( 3x+2)( x-1)}{2} =\frac{3x^{2} -x-2}{2} \end{array}$
