Question

Помогите с алгеброй пожалуйста

Answer 1

Объяснение:

1.

$(2b+5)^2-(b-3)(3b+5)=4b^2+20b+25-(3b^2-4b-15)=\\=4b^2+20b+25-3b^2+4b+15=b^2+24b+40.$

2.

$1)\ 6a^2b^2-600a^2c^2=6*(a^2b^2-100a^2c^2)=6*((ab)^2-(10ac)^2)=\\=6*(ab+10ac)*(ab-10ac).\\2)\ 7a^2-28ab+28b^2=7*(a^2-4ab+4b^2)=7*(a^2-4*a*2b+(2b)^2)=\\=7*(a-2b)^2.$

3.

$(0;-36)\ \ \ \ \ (4;0)\ \ \ \ y=kx+b\ ?\\ \left \{ {{-36=k*0+b} \atop {0=k*4+b}} \right. \ \ \ \ \ \left \{ {{b=-36} \atop {4k-36=0}\right. \ \ \ \ \ \left \{ {{b=-36} \atop {4k=36\ |:4}} \right.\ \ \ \ \ \left \{ {{b=-36} \atop {k=9}} \right. .$

Ответ: k=9,   b=-36.