Question

Ajutor clasa8 va rog b) nu 

Answer 1

$a)$$3(2a-5b) ^{2} -12(a-b) ^{2} =3(4a ^{2} -20ab+25b ^{2}) -12(a ^{2} -2ab+b ^{2} )=$$12a ^{2} -60ab+75b ^{2} -12a ^{2} +24ab-12b ^{2} =-36ab+63b ^{2}=$$9b(-4a+7b)$

$c)$$(3 x^{2} +4) ^{2} +(3 x^{2} -4) ^{2} -2(5-3 x^{2} )(5+3 x^{2} )=$$9 x^{4} +24 x^{2} +16+9 x^{4} -24 x^{2} +16-2(25-9 x^{4} )=$$18 x^{4} +32-50+18 x^{4} =36 x^{4} -18=18(2 x^{4} -1)$

$d)$$(4 a^{3} +5) ^{2} +(4 a^{3} -1) ^{2}-2(4 a^{3} +5) (4 a^{3}- 1)$$=16a x^{6} +40a ^{3} +25+16a x^{6} -8a ^{3} +1-2(16a ^{6} -4a ^{3} +20a^{3}-5)=$$32a ^{6} +32a ^{3}+26-2(16a x^{6} +16a ^{3} -5)=$$32a ^{6} +32a ^{3}+26-32a ^{6} -32a ^{3}+10 =36$

$e)$$(p-2a)(p+2a)-(p-a)(p ^{2} +pa+a ^{2} )$=$p ^{2} -4a ^{2} -(p ^{3} +p ^{2} a-a ^{3} -p ^{2} a-pa ^{2} -a ^{3} )=p ^{2} -4a ^{2}-(p ^{3} -a ^{3})=$$p ^{2} -4a ^{2}-(p ^{3} -a ^{3})=p ^{2} -4a ^{2}-p ^{3} +a ^{3}=p ^{2} (1-p)+a ^{2} (a-4)$

$e)$$x(2x-1) ^{2} +2(x+1)(x^{2} -x+1)=$$x(4 x^{2} -4x+1)+2( x^{3} - x^{2} +x+ x^{2} -x+1)=$$=6x^{3} -4 x^{2} +x+2=x(6x ^{2} -4x+1)+2$