2 варианта:)и оба задания помогите:))

Question

Дан 1 ответ

мохинсан_zn · Answer 1 · 2018-03-17T17:35:42+0000

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)
$5y^2\cdot \frac{1}{2}y+ \frac{1}{4}y\cdot8y+3y-2y\cdot \frac{1}{6}y -0,5y\cdot y^2+1=\\ =(5\cdot \frac{1}{2}-0,5)y^{2+1}+( \frac{1}{4}\cdot \frac{8}{1}- \frac{2}{1}\cdot \frac{1}{6})y^{1+1}+3y+1=\\ =(5,5-0,5)y^3+(2- \frac{1}{3})y^2+3y+1=\\ =5y^3+( \frac{6-1}{3})y^2+7y+1=5y^3+ \frac{5}{3}y^2+7y+1=5y^3+1 \frac{2}{3}y^2+7y+1$
2.
\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+3\cdot(-2)^2+2\cdot(-2)\cdot(-1)\}=\\ =1\cdot\{1+4+4)=1+4+4=5" alt="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+3\cdot(-2)^2+2\cdot(-2)\cdot(-1)\}=\\ =1\cdot\{1+4+4)=1+4+4=5" align="absmiddle" class="latex-formula">

Вариант 4
1.
а)
$10b^2+b^3-4b^2+1-3b+5=b^3+6b^2-3b+6;$
б)
$\frac{3}{2}t\cdot \frac{7}{3}t^2-8t\cdot \frac{3}{4}t+ \frac{4}{7}t-t\cdot2t-t\cdot3,5t^2-2=\\ = ( \frac{3}{2}\cdot \frac{7}{3}-3,5)t^3-( \frac{8}{1}\cdot \frac{3}{4}+2)t^2+ \frac{4}{7}t-2=\\ =( \frac{7}{2}-3,5)t^3-(6+2)t^2+\frac{4}{7}t-2=(3,5-3,5)t^3-8t^2+ \frac{4}{7}t-2=\\ =8t^2+ \frac{4}{7}t-2;$
2.
\left \{ {{x=-3} \atop {y=1}} \right. \\ p(-3;1)=2\cdot1^4+4\cdot(-3)\cdot1(-3+1^2)=\\ =2-12\cdot(-2)=2+24=26" alt="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)=2\cdot1^4+4\cdot(-3)\cdot1(-3+1^2)=\\ =2-12\cdot(-2)=2+24=26" align="absmiddle" class="latex-formula">