Question

д)е)ж)з) помогитепжж

Answer 1

Ответ:

Объяснение:

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Answer 2

Ответ:

д)

$(2a+3b)^2-(3a+2b)^2=\Big((2a+3b)-(3a+2b)\Big)\Big((2a+3b)+(3a+2b)\Big)=\\\\=(b-a)(5a+5b)=5(b-a)(a+b)=5(b^2-a^2)=5b^2-5a^2$

е)

$2(3x+y)^2-3(2x+3y)^2=\\\\=\Big(\sqrt2(3x+y)-\sqrt3(2x+3y)\Big)\Big(\sqrt2(3x+y)+\sqrt3(2x+3y)\Big)=\\\\=\Big((3\sqrt2-2\sqrt3)x+(\sqrt2-3)y\Big)\Big((3\sqrt2+2\sqrt3)x+(\sqrt2+3)y\Big)\\\\ili\\\\2(3x+y)^2-3(2x+3y)^2=2(9x^2+6xy+y^2)-3(4x^2+12xy+9y^2)=\\\\=6x^2-24xy-25y^2$

ж)

$(m+n)^2+2(m+n)(2m-n)+(2m-n)^2=\\\\=\Big((m+n)+(2m-n)\Big)^2=\Big(3m\Big)^2=9m^2$

з)

$2(p+3q)(p+2q)-(p+2q)^2-(3q+p)^2=\\\\=-\Big((p+3q)^2-2(p+3q)(p+2q)+(p+2q)^2\Big)=\\\\=-\Big((p+3q)-(p+2q)\Big)^2=-\Big(p-p+3q-2q\Big)^2=-q^2$