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

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