Упр.30.13 ГДЗ Мордкович 7 класс (Алгебра)

а) 6а2 — (2 — (1,56а — (а2 + 0,36а)) + (5,5а2 + 1,2а — 1));
б) (а2 + 2×2) — (5а2 — 1,2аx + (2,8×2 — (1,5а2 — 0,5аx + 1,8×2)));
в) 12,5×2 + у2 — (8×2 — 5у2 — (-10×2 + (5,5×2 — 6у2)));
г) (у3 + 3z2) — (у3 — баz + (2у3 — (3z2 + 4аz — 1,2у3))).

1. а)

   $$6a^2-(2-(1{,}56a-(a^2+0{,}36a))+(5{,}5a^2+1{,}2a-1))=$$
   $$=6a^2-(2-(1{,}56a-a^2-0{,}36a)+5{,}5a^2+1{,}2a-1)$$
   $$=6a^2-(2-(1{,}2a-a^2)+5{,}5a^2+1{,}2a-1)$$
   $$=6a^2-(2-1{,}2a+a^2+5{,}5a^2+1{,}2a-1)$$
   $$=6a^2-(1+6{,}5a^2)$$
   $$=6a^2-1-6{,}5a^2=-0{,}5a^2-1.$$

2. б)

   $$(a^2+2x^2)-(5a^2-1{,}2ax+(2{,}8x^2-(1{,}5a^2-0{,}5ax+1{,}8x^2)))=$$
   $$=a^2+2x^2-(5a^2-1{,}2ax+(2{,}8x^2-1{,}5a^2+0{,}5ax-1{,}8x^2))$$
   $$=a^2+2x^2-(5a^2-1{,}2ax+(x^2-1{,}5a^2+0{,}5ax))$$
   $$=a^2+2x^2-(5a^2-1{,}2ax+x^2-1{,}5a^2+0{,}5ax)$$
   $$=a^2+2x^2-(3{,}5a^2-0{,}7ax+x^2)$$
   $$=a^2+2x^2-3{,}5a^2+0{,}7ax-x^2$$
   $$=-2{,}5a^2+x^2+0{,}7ax.$$

3. в)

   $$12{,}5x^2+y^2-(8x^2-5y^2-(-10x^2+(5{,}5x^2-6y^2)))=$$
   $$=12{,}5x^2+y^2-(8x^2-5y^2-(-10x^2+5{,}5x^2-6y^2))$$
   $$=12{,}5x^2+y^2-(8x^2-5y^2-(-4{,}5x^2-6y^2))$$
   $$=12{,}5x^2+y^2-(8x^2-5y^2+4{,}5x^2+6y^2)$$
   $$=12{,}5x^2+y^2-(12{,}5x^2+y^2)=0.$$

4. 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) $$-3{,}2y^3+6z^2+10az$$.