Упр.30.13 ГДЗ Мордкович 7 класс (Алгебра)
- Преобразуйте выражение в многочлен стандартного вида:
а) $$6a^2-\left(2-\left(1{,}56a-\left(a^2+0{,}36a\right)\right)+\left(5{,}5a^2+1{,}2a-1\right)\right);$$
б) $$\left(a^2+2x^2\right)-\left(5a^2-1{,}2ax+\left(2{,}8x^2-\left(1{,}5a^2-0{,}5ax+1{,}8x^2\right)\right)\right);$$
в) $$12{,}5x^2+y^2-\left(8x^2-5y^2-\left(-10x^2+\left(5{,}5x^2-6y^2\right)\right)\right);$$
г) $$\left(y^3+3z^2\right)-\left(y^3-6az+\left(2y^3-\left(3z^2+4az-1{,}2y^3\right)\right)\right).$$
а)
$$6a^2-(2-(1{,}56a-(a^2+0{,}36a))+(5{,}5a^2+1{,}2a-1))=$$
$$=6a^2-(2-(1{,}56a-a^2-0{,}36a)+5{,}5a^2+1{,}2a-1)$$
$$=6a^2-(2-(1{,}2a-a^2)+5{,}5a^2+1{,}2a-1)$$
$$=6a^2-(2-1{,}2a+a^2+5{,}5a^2+1{,}2a-1)$$
$$=6a^2-(1+6{,}5a^2)$$
$$=6a^2-1-6{,}5a^2=-0{,}5a^2-1.$$б)
$$(a^2+2x^2)-(5a^2-1{,}2ax+(2{,}8x^2-(1{,}5a^2-0{,}5ax+1{,}8x^2)))=$$
$$=a^2+2x^2-(5a^2-1{,}2ax+(2{,}8x^2-1{,}5a^2+0{,}5ax-1{,}8x^2))$$
$$=a^2+2x^2-(5a^2-1{,}2ax+(x^2-1{,}5a^2+0{,}5ax))$$
$$=a^2+2x^2-(5a^2-1{,}2ax+x^2-1{,}5a^2+0{,}5ax)$$
$$=a^2+2x^2-(3{,}5a^2-0{,}7ax+x^2)$$
$$=a^2+2x^2-3{,}5a^2+0{,}7ax-x^2$$
$$=-2{,}5a^2+x^2+0{,}7ax.$$в)
$$12{,}5x^2+y^2-(8x^2-5y^2-(-10x^2+(5{,}5x^2-6y^2)))=$$
$$=12{,}5x^2+y^2-(8x^2-5y^2-(-10x^2+5{,}5x^2-6y^2))$$
$$=12{,}5x^2+y^2-(8x^2-5y^2-(-4{,}5x^2-6y^2))$$
$$=12{,}5x^2+y^2-(8x^2-5y^2+4{,}5x^2+6y^2)$$
$$=12{,}5x^2+y^2-(12{,}5x^2+y^2)=0.$$г)
$$(y^3+3z^2)-(y^3-6az+(2y^3-(3z^2+4az-1{,}2y^3)))=$$
$$=y^3+3z^2-(y^3-6az+(2y^3-3z^2-4az+1{,}2y^3))$$
$$=y^3+3z^2-(y^3-6az+(3{,}2y^3-3z^2-4az))$$
$$=y^3+3z^2-(4{,}2y^3-10az-3z^2)$$
$$=y^3+3z^2-4{,}2y^3+10az+3z^2$$
$$=-3{,}2y^3+6z^2+10az.$$
Ответ
а) $$-0{,}5a^2-1$$; б) $$-2{,}5a^2+x^2+0{,}7ax$$; в) $$0$$; г) $$-3{,}2y^3+6z^2+10az$$.
