Question

2 варианта:)и оба задания помогите:))

Answer 1

Вариант3
1.
а)
$9c^4+2c-5c^2+7-2c-c^2=\ =9c^4-5c^2-c^2+2c-2c+7=9c^4-6c^2+7;$
б)
$5y^2cdot frac{1}{2}y+ frac{1}{4}ycdot8y+3y-2ycdot frac{1}{6}y -0,5ycdot y^2+1=\ =(5cdot frac{1}{2}-0,5)y^{2+1}+( frac{1}{4}cdot frac{8}{1}- frac{2}{1}cdot frac{1}{6})y^{1+1}+3y+1=\ =(5,5-0,5)y^3+(2- frac{1}{3})y^2+3y+1=\ =5y^3+( frac{6-1}{3})y^2+7y+1=5y^3+ frac{5}{3}y^2+7y+1=5y^3+1 frac{2}{3}y^2+7y+1$
2.
$p(-2,-1)-?\ p(-2,-1)==> left { {{x=-2} atop {y=-1}} right. \ p(x,y)=x^3-3x^2y+xy^2-y^3+6x^2y+xy^2-x^3=\ =x^3-x^3-y^3+(6-3)x^2y+(1+1)xy^2=\ =-y(1+3x^2+2xy);\ p(-2,-1)=-(-1)cdot{1+3cdot(-2)^2+2cdot(-2)cdot(-1)}=\ =1cdot{1+4+4)=1+4+4=5$



Вариант 4
1.
а)
$10b^2+b^3-4b^2+1-3b+5=b^3+6b^2-3b+6;$
б)
$frac{3}{2}tcdot frac{7}{3}t^2-8tcdot frac{3}{4}t+ frac{4}{7}t-tcdot2t-tcdot3,5t^2-2=\ = ( frac{3}{2}cdot frac{7}{3}-3,5)t^3-( frac{8}{1}cdot frac{3}{4}+2)t^2+ frac{4}{7}t-2=\ =( frac{7}{2}-3,5)t^3-(6+2)t^2+frac{4}{7}t-2=(3,5-3,5)t^3-8t^2+ frac{4}{7}t-2=\ =8t^2+ frac{4}{7}t-2;$
2.
$p(-3;1)-?\ p(x,y)=x^4+4x^2y-6x^2y^2+4xy^3-x^4+y^4+6x^2y^2+y^4=\ =x^4-x^4-6x^2y^2+6x^2y^2+y^4+y^4+4x^2y+4xy^3=\ =2y^4+4xy(x+y^2);\ p(-3,1)=> left { {{x=-3} atop {y=1}} right. \ p(-3;1)=2cdot1^4+4cdot(-3)cdot1(-3+1^2)=\ =2-12cdot(-2)=2+24=26$