Плачу много баллов, решите пожалуйста.

Question 1

Question 2

1.
а) $(x-2y)(x+2y)+4y^{2}=x^{2}-4y^{2}+4y^{2}=x^{2}$
в) $(5x-1)^{2}+10x=25x^{2}+1-10x+10x=25x^{2}+1$
д) $(m-2n)(m^{2}+2mn+4n^{2})+6n^{3}=m^{3}-8n^{3}+6n^{3}=m^{3}-2n^{3}$
ж) $(3x-4y)^{2}-(2x-7y)(4x+2y)=9x^{2}+16y^{2}-24xy-8x^{2}+28xy- \\ -4xy+14y^{2}=x^{2}+30y^{2}$

2.
в) $8a^{3}+c^{3}=(2a+c)(4a^{2}-2ac+c^{2})$
д) $9ab^{2}-16ac^{2}=a(9b^{2}-16c^{2})=a(3b-4c)(3b+4c)$

3.
$\frac{x(x+6)+2(x+8)}{(x+8)^{2}-16}= \frac{x^{2}+6x+2x-16}{x^{2}+64+16x-16}= \frac{x^{2}+8x-16}{x^{2}+48+16x}$
$\frac{xy^{3}+x^{4}}{y^{5}-4x^{4}y}*( \frac{x+y}{x-y}- \frac{x^{2}-xy+y^{2}}{x^{2}+xy+y^{2}})= \frac{x(y^{3}+x^{3})}{y(y^{4}-4x^{4})}*(\frac{(x+y)(x^{2}+xy+y^{2})}{(x-y)(x^{2}+xy+y^{2})}-\\-\frac{(x-y)(x^{2}-xy+y^{2})}{(x-y)(x^{2}+xy+y^{2})})$
$\frac{x+3}{x^{2}+9} ( \frac{x+3}{x-3}+ \frac{x-3}{x+3} )= \frac{1}{x-3} * \frac{(x+3)^{2}+(x-3)^{2}}{(x+3)(x-3)}= \frac{1}{x-3} * \frac{x^{2}+9-6x+x^{2}+9-6x}{(x-3)(x+3)} = \\ =\frac{1}{x-3} * \frac{2x^{2}+18-12x}{(x-3)(x+3)} =\frac{2(x^{2}+9-6x)}{(x-3)(x-3)(x+3)}= \frac{2}{x+3}$

4.
1) $(15a-12x)^{2}=...+144x^{2}=225a^{2}-360ax+144x^{2}$
2) $(4+3xy)^{2}=16+24xy+9x^{2}y^{2}$
3) $(5-b^{2})(b^{2}+5)=25-b^{4}$
4) $(17-3a)(17+3a)=289-9a^{2}$
5) $x^{3}+0,125=(x+0,5)(x^{2}-0,5x+0,25)$
6) $\frac{18a^{3}b^{4}}{12ab^{2}}= \frac{3a^{2}b^{2}}{2}$
7) $\frac{a^{2}-4}{3a+6} = \frac{a-2}{3}$
8) $12x^{2}y^{4}: \frac{4y^{4}}{x} =3x^{3}$