Упр.1092 ГДЗ Алимов 10-11 класс (Алгебра)

1) (1+x)8;
2) (x+1)7;
3) (a-1)9;
4) (y-1)10;
5) (2x+1)5;
6) (x+2)6;
7) (3x+2)4;
8) (2a+3)5;
9) (2a — 1/2)5;
10) (3x — 1/3)4.

Используем формулу бинома Ньютона:

$$ (a+b)^n=C_n^0a^n+C_n^1a^{n-1}b+C_n^2a^{n-2}b^2+\dots+C_n^nb^n $$

Тогда получаем:

$$ (1+x)^8=1+8x+28x^2+56x^3+70x^4+56x^5+28x^6+8x^7+x^8 $$

$$ (x+1)^7=x^7+7x^6+21x^5+35x^4+35x^3+21x^2+7x+1 $$

$$ (a-1)^9=a^9-9a^8+36a^7-84a^6+126a^5-126a^4+84a^3-36a^2+9a-1 $$

$$ (y-1)^{10}=y^{10}-10y^9+45y^8-120y^7+210y^6-252y^5+210y^4-120y^3+45y^2-10y+1 $$

$$ (2x+1)^5=32x^5+80x^4+80x^3+40x^2+10x+1 $$

$$ (x+2)^6=x^6+12x^5+60x^4+160x^3+240x^2+192x+64 $$

$$ (3x+2)^4=81x^4+216x^3+216x^2+96x+16 $$

$$ (2a+3)^5=32a^5+240a^4+720a^3+1080a^2+810a+243 $$

$$ \left(2a-\frac12\right)^5=32a^5-40a^4+20a^3-5a^2+\frac58a-\frac1{32} $$

$$ \left(3x-\frac13\right)^4=81x^4-36x^3+6x^2-\frac49x+\frac1{81} $$

Ответ

1) $$1+8x+28x^2+56x^3+70x^4+56x^5+28x^6+8x^7+x^8$$
2) $$x^7+7x^6+21x^5+35x^4+35x^3+21x^2+7x+1$$
3) $$a^9-9a^8+36a^7-84a^6+126a^5-126a^4+84a^3-36a^2+9a-1$$
4) $$y^{10}-10y^9+45y^8-120y^7+210y^6-252y^5+210y^4-120y^3+45y^2-10y+1$$
5) $$32x^5+80x^4+80x^3+40x^2+10x+1$$
6) $$x^6+12x^5+60x^4+160x^3+240x^2+192x+64$$
7) $$81x^4+216x^3+216x^2+96x+16$$
8) $$32a^5+240a^4+720a^3+1080a^2+810a+243$$
9) $$32a^5-40a^4+20a^3-5a^2+\frac58a-\frac1{32}$$
10) $$81x^4-36x^3+6x^2-\frac49x+\frac1{81}$$