Đáp án+Giải thích các bước giải:

$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ toán\ 1:\\ 1.\ ( x+2y)^{2} =x^{2} +4xy+4y^{2}\\ 2.\ ( 2x+3y)^{2} =4x^{2} +12xy+9y^{2}\\ 3.\ ( 3x-2y)^{2} =9x^{2} -12xy+4y^{2}\\ 4.\ ( 5x-y)^{2} =25x^{2} -10xy+y^{2}\\ 5.\ \left( x+\frac{1}{4}\right)^{2} =x^{2} +\frac{1}{2} x+\frac{1}{16}\\ 6.\ \left( 2x-\frac{1}{2}\right)^{2} =4x^{2} -2x+\frac{1}{4}\\ 7.\ \left(\frac{1}{3} x-\frac{1}{2} y\right)^{2} =\frac{1}{9} x-\frac{1}{3} xy-\frac{1}{4} y^{2}\\ 8.\ ( 3x+1)( 3x-1) =9x^{2} -1\\ 9.\ \left( x^{2} +\frac{2}{5} y\right)\left( x^{2} -\frac{2}{5} y\right) =x^{4} -\frac{4}{25} y^{2}\\ 10.\ \left(\frac{x}{2} -y\right)\left(\frac{x}{2} +y\right) =\frac{x^{2}}{4} -y^{2}\\ 11.\ \left(\frac{x}{2} -2y\right)^{2} =\frac{x}{4} -2xy+4y^{2}\\ 12.\ \left(\sqrt{2} x-y\right)^{2} =2x-2\sqrt{2} xy+y^{2}\\ 13.\ \left(\frac{3}{2} x+3y\right)^{2} =\frac{9}{4} x^{2} +9xy+9y^{2}\\ 14.\ \left(\sqrt{2} x+\sqrt{8} y\right)^{2} =2x+2\sqrt{16} xy+8y^{2} =2x+8xy+8y^{2}\\ 15.\ \left( x+\frac{1}{6} y+3\right)^{2} =x^{2} +\frac{y^{2}}{36} +9+2x\frac{1}{6} y+2x.3+\frac{1}{6} y.3\\ =x^{2} +\frac{y^{2}}{36} +9+\frac{1}{3} xy+6x+\frac{1}{2} y\\ 16.\ \left(\frac{1}{2} x-4y\right)^{2} =\frac{x^{2}}{4} -4xy+16y^{2}\\ 17.\ \left(\frac{x}{2} +2y^{2}\right)\left(\frac{x}{2} -2y^{2}\right) =\frac{x^{2}}{4} -4y^{4}\\ 18.\ \left( x^{2} -4\right)\left( x^{2} +4\right) =x^{4} -16\\ 19.\ ( x+y)^{2} +( x-y)^{2} =x^{2} +2xy+y^{2} +x^{2} -2xy+y^{2} =2x^{2} +2y^{2}\\ 20.\ ( 2x+3)^{2} -( x+1)^{2} =4x^{2} +12x+9-\left( x^{2} +2x+1\right) =3x^{2} +10x+8\\ Bài\ toán\ 2:\\ Cột\ 1:\\ 1.\ \left( x+\frac{1}{3}\right)^{3} =x^{3} +3x^{2} .\frac{1}{3} +3x.\left(\frac{1}{3}\right)^{2} +\left(\frac{1}{3}\right)^{3} =x^{3} +x^{2} +\frac{1}{3} x+\frac{1}{27}\\ 2.\ \left( 2x+y^{2}\right)^{3} =( 2x)^{3} +3.( 2x)^{2} .y^{2} +3.2x.\left( y^{2}\right)^{2} +\left( y^{2}\right)^{3}\\ =8x^{3} +12x^{2} y^{2} +6xy^{4} +y^{6}\\ 3.\ \left(\frac{1}{2} x^{2} +\frac{1}{3} y\right)^{3} =\left(\frac{1}{2} x^{2}\right)^{3} +3.\left(\frac{1}{2} x^{2}\right)^{2} .\frac{1}{3} y+3.\frac{1}{2} x.\left(\frac{1}{3} y\right)^{2} +\left(\frac{1}{3} y\right)^{3}\\ =\frac{x^{3}}{8} +\frac{x^{3} y}{4} +\frac{xy^{2}}{6} +\frac{y^{3}}{27}\\ 4.\ \left( 3x^{2} -2y\right)^{3} =\left( 3x^{2}\right)^{3} -3.\left( 3x^{2}\right)^{2} .2y+3.3x^{2} .( 3y)^{2} -( 2y)^{3}\\ =27x^{6} -54x^{4} y+81x^{2} y^{2} -8y^{3}\\ 5.\ \left(\frac{2}{3} x^{2} -\frac{1}{2} y\right)^{3} =\left(\frac{2}{3} x^{2}\right)^{3} -3.\left(\frac{2}{3} x^{2}\right)^{2} .\frac{1}{2} y+3.\frac{2}{3} x^{2} .\left(\frac{1}{2} y\right)^{2} -\left(\frac{1}{2} y\right)^{3}\\ =\frac{8}{27} x^{6} -\frac{2}{3} x^{4} y+\frac{1}{2} x^{3} y^{2} -\frac{1}{8} y^{3}\\ 6.\ \left( 2x+\frac{1}{2}\right)^{3} =( 2x)^{3} +3.( 2x)^{2} .\frac{1}{2} +3.2x.\left(\frac{1}{2}\right)^{2} +\left(\frac{1}{2}\right)^{3}\\ =8x^{3} +6x^{2} +\frac{3}{2} x+\frac{1}{8}\\ 7.\ ( x-3)^{2} =x^{3} -3.x^{2} .3+3.x.3^{2} -3^{3} =x^{3} -9x^{2} +27x-27\\ Cột\ 2:\\ 1.\ ( x+1)\left( x^{2} -x+1\right) =x^{3} +1\\ 2.\ ( x-3)\left( x^{2} +3x+9\right) =x^{3} -3^{3} =x^{3} -27\\ 3.\ ( x-2)\left( x^{2} +2x+4\right) =x^{3} -2^{3} =x^{3} -8\\ 4.\ ( x+4)\left( x^{2} -4x+16\right) =x^{3} +4^{3} =x^{3} +64\\ 5.\ ( x-3y)\left( x^{2} +3xy+9y^{2}\right) =x^{3} -( 3y)^{3} =x^{3} -27y^{3}\\ 6.\ \left( x^{2} -\frac{1}{3}\right)\left( x^{4} +\frac{1}{3} x^{2} +\frac{1}{9}\right) =\left( x^{2}\right)^{3} -\left(\frac{1}{3}\right)^{3} =x^{6} -\frac{1}{27}\\ 7.\ \left(\frac{1}{3} x+2y\right)\left(\frac{1}{9} x^{2} -\frac{2}{3} xy+4y^{2}\right) =\left(\frac{1}{3} x\right)^{3} +( 2y)^{3} =\frac{x^{3}}{27} +8y^{3} \end{array}$