Упр.489 ГДЗ Колягин Ткачёва 11 класс (Алгебра)

1) (2a-1)5;
2) (1/2+2b)6;
3) (3x + 1/3)4;
4) (y/3 — 3)5.

1. $$\left(2a-1\right)^5=\sum_{k=0}^{5} C_5^k (2a)^{5-k}(-1)^k$$
   $$\left(2a-1\right)^5=32a^5-80a^4+80a^3-40a^2+10a-1.$$
2. $$\left(\frac12+2b\right)^6=\sum_{k=0}^{6} C_6^k \left(\frac12\right)^{6-k}(2b)^k$$
   $$\left(\frac12+2b\right)^6=\frac1{64}+\frac38b+\frac{15}{4}b^2+20b^3+60b^4+96b^5+64b^6.$$
3. $$\left(3x+\frac13\right)^4=\sum_{k=0}^{4} C_4^k (3x)^{4-k}\left(\frac13\right)^k$$
   $$\left(3x+\frac13\right)^4=81x^4+36x^3+6x^2+\frac49x+\frac1{81}.$$
4. $$\left(\frac{y}{3}-3\right)^5=\sum_{k=0}^{5} C_5^k \left(\frac{y}{3}\right)^{5-k}(-3)^k$$
   $$\left(\frac{y}{3}-3\right)^5=\frac1{243}y^5-\frac5{27}y^4+\frac{10}{3}y^3-30y^2+135y-243.$$

Ответ

1) $$32a^5-80a^4+80a^3-40a^2+10a-1$$
2) $$\frac1{64}+\frac38b+\frac{15}{4}b^2+20b^3+60b^4+96b^5+64b^6$$
3) $$81x^4+36x^3+6x^2+\frac49x+\frac1{81}$$
4) $$\frac1{243}y^5-\frac5{27}y^4+\frac{10}{3}y^3-30y^2+135y-243$$