Bài 3.
5. $1-x+\frac{1}{4}x^2=1^2-2.1.\frac{1}{2}x+(\frac{1}{2}x)^2$
$=(1-\frac{1}{2}x)^2$
6. $-xy+\frac{x^2}{4}+y^2=(\frac{x}{2})^2-2.\frac{x}{2}.y+y^2$
$=(\frac{x}{2}-y)^2$
7. Tương tự câu 6 thì $\frac{y^2}{4}+x^2-xy=(x-\frac{y}{2})^2$
10. $16x^2-8xy+y^2=(4x)^2-2.4x.y+y^2$
$=(4x-y)^2$
14.
$-2xy+\frac{81x^2}{49}+\frac{49y^2}{81}$
$=(\frac{9}{7}x)^2+(\frac{7}{9}y)^2-2.\frac{9}{7}x.\frac{7}{9}y$
$=(\frac{9}{7}x-\frac{7}{9}y)^2$
16.
$-3x+x^2+\frac{9}{4}=x^2-2.x.\frac{3}{2}+(\frac{3}{2})^2$
$=(x-\frac{3}{2})^2$
17.
$(25x^2-10xy+y^2)-(5x-y)+\frac{1}{4}$
$=(5x-y)^2-2.(5x-y).\frac{1}{2}+(\frac{1}{2})^2$
$=(5x-y-\frac{1}{2})^2$
18.
$(4x^2+2x+\frac{1}{4})-2(2x+\frac{1}{2})+1$
$=(2x+\frac{1}{2})^2-2(2x+\frac{1}{2}).1+1^2$
$=(2x+\frac{1}{2}-1)^2=(2x-\frac{1}{2})^2$
Bài 4:
1. $(1-\frac{m}{2})^2=1-m+\frac{m^2}{4}$
2. $(6x-5)^2=(6x)^2-2.6x.5+5^2=36x^2-60x+25$
3. $(\frac{x}{10}-5y)^2=(\frac{x}{10})^2-2.\frac{x}{10}.5y+(5y)^2$
$=\frac{x^2}{100}-xy+25y^2$
4. $(3x^2-5y^2)^2=(3x^2)^2-2.3x^2.5y^2+(5y^2)^2$
$=9x^4-30x^2y^2+25y^4$
5. $(\frac{2}{3}m-3n)^2=(\frac{2}{3}m)^2-2.\frac{2}{3}m.3n+(3n)^2$
$=\frac{4}{9}m^2-4mn+9n^2$
6. $(\frac{5}{6}n-\frac{3}{5}m)^2=(\frac{5}{6}n)^2-2.\frac{5}{6}n.\frac{3}{5}m+(\frac{3}{5}m)^2$
$=\frac{25}{36}n^2-nm+\frac{9}{25}m^2$
