Question

Помогите пж алгебра 7класс

Answer 1

Ответ:

Объяснение:

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Answer 2

Ответ:

$1a)\ \ a=-3,5\ \ ,\ \ (a+3)^2-(a-2)(a+2)=\\\\=a^2+6a+9-(a^2-4)=6a+13=-6\cdot 3,5+13=-8\\\\b)\ \ x=-0,1\ \ ,\ \ \ (x-3)^2-(x+3)(x-3)=\\\\=x^2-6x+9-(x^2-9)=-6x+18=6\cdot 0,1+18=18,6\\\\c)\ \ m=-0,5\ \ ,\ \ \ (m+3)^2-(m-9)(m+9)=\\\\=m^2+6m+9-(m^2-81)=6m+90=-6\cdot 0,5+90=87\\\\d)\ \ c=\dfrac{1}{4}\ \ ,\ \ \ (c+2)^2-(c+4)(c-4)=\\\\=c^2+4c+4-(c^2-16)=4c+20=4\cdot \dfrac{1}{4}+20=21$

$42a)\ \ a=-6\ \ ,\ \ \ (5a-10)^2-(3a-8)^2+132a=\\\\=25a^2-100a+100-(9a^2-48a+64)+132a=16a^2+80a+36=\\\\=4\, (4a^2+20a+9)=4\, (144-120+9)=4\cdot 33=132\\\\b)\ \ p=-2\ \ ,\ \ \ (3p-8)^2+(4p+6)^2+100p=\\\\=9p^2-48p+64+(16p^2+48p+36)+100p=25p^2+100p+100=\\\\=25\, (p^2+4p+4)=25\, (p+2)^2=25\, (-2+2)^2=0\\\\c)\ \ b=-1\ \ ,\ \ \ (5b-3)^2+(12b-4)^2-4b=\\\\=25b^2-30b+9+(144b^2-96b+16)-4b=169b^2-130b+25=\\\\=(13b-5)^2=(-13-5)^2=(-18)^2=324$

$d)\ \ m=-\dfrac{2}{3}\ \ ,\ \ \ (13-5m)^2-(12-4m)^2+4m=\\\\=169-130m+25m^2-(144-96m+16m^2)+4m=9m^2-30m+25=\\\\=(3m-5)^2=(-2-5)^2=(-7)^2=49$