Đáp án:
Giải thích các bước giải :
c.1 $( \frac{1}{2} - x )^{2}$
= ($\frac{1}{2}$ - x)($\frac{1}{2}$ -x)
= $\frac{1}{2}$ ($\frac{1}{2}$ - x ) - x ( $\frac{1}{2}$ - x )
= $\frac{1}{4}$ - $\frac{1}{2}$x - $\frac{1}{2}$x + $x^{2}$
= ($\frac{1}{4}$ -x )+ $x^{2}$
c.2) $(2x-1)^{2}$
= (2x-1)(2x-1)
= 2x(2x-1)-(2x-1)
= 4$x^{2}$ - 2x - 2x + 1
= ( 4$x^{2}$ - 4x )+ 1
d.1)$(2x-3y)^{2}$
= (2x-3y)(2x-3y)
= 2x(2x-3y) - 3y(2x-3y)
= 4$x^{2}$ - 6xy - 6xy + 9$y^{2}$
=( 4$x^{2}$ - 12xy )+ 9$y^{2}$
d.2)$(0,01-xy)^{2}$
= (0,01 - xy) ( 0,01 - xy )
= 0,01(0,01-xy)-xy(0,01-xy)
= 0,0001 - 0,01xy - 0,01xy + $x^{2}$ $y^{2}$
=( 0,0001 - 0,02xy )+ $x^{2}$ $y^{2}$
g) (x+y+z).(x-y-z)
= x(x-y-z) + y(x-y-z) + z(x-y-z)
= $x^{2}$ - xy - xz + xy - $y^{2}$ - yz + xz - yz - $z^{2}$
= $x^{2}$ - $y^{2}$ - $z^{2}$
h) (x-y+z) ( x+y+z )
= x (x-y+z) + y(x-y+z) + z(x-y+z)
= $x^{2}$ - xy + xz + xy - $y^{2}$ + yz + xz - yz + $z^{2}$
= $x^{2}$ - $y^{2}$ + $z^{2}$ + 2xz
