1)
1. (3/5)*a³ⁿ⁺¹*b²+(2/3)*aⁿ⁻¹*b³)²=
=(9/25)*a⁶ⁿ⁺²*b₄+2*(3/5)a³ⁿ⁺¹*b²*(2/3)aⁿ⁻¹*b³+(4/9)*a²ⁿ⁻²*b⁶=
=(9/25)*a⁶ⁿ⁺²b⁴+(4/5)*a⁴ⁿ*b⁵+(4/9)*a²ⁿ⁻²*b⁶.
2. (4/45)*a²ⁿ⁻²b⁵*(9*a²ⁿ⁺²+5b)=(4/5)*a⁴ⁿ*b⁵+(4/9)*a²ⁿ⁻²*b⁶.
3. (9/25)*a⁶ⁿ⁺²*b⁴+(4/5)*a⁴ⁿ*b⁵+(4/9)*a²ⁿ⁻²*b⁶-(4/5)*a⁴*b⁵-(4/9)*a²ⁿ⁻²*b⁶+ +(16/25)*a⁶ⁿ⁺²*b⁴==(9/25)*a⁶ⁿ⁺²+(4/5)*a⁴ⁿ*b⁵-(4/5)*a⁴ⁿb⁵+(16/25)*a⁶ⁿ⁺²*b⁴ a⁶ⁿ⁺²*b⁴≡a⁶ⁿ⁺²*b⁴.
2)
1. (5/6)*x²ⁿ⁻¹*yⁿ-(3/5)*xⁿ⁺¹*y²)²=(25/36)*x⁴ⁿ⁻²*y²ⁿ-2*(5/6)*x²ⁿ⁻¹*yⁿ*(3/5)*xⁿ⁺¹*y²+
+(9/25)x²ⁿ⁺²*y⁴=(25/36)*x⁴ⁿ⁻²*y²ⁿ-x³ⁿ*yⁿ⁺²+(9/25)*x²ⁿ⁺²*y⁴.
2. (1/36)*x³ⁿ*yⁿ⁺²*(25xⁿ⁻²*yⁿ⁻²-36)=(25/36)*x⁴ⁿ⁻²*y²ⁿ⁺²-x³ⁿ*yⁿ⁺².
(25/36)x⁴ⁿ⁻²*y²ⁿ-x³ⁿ*yⁿ⁺²+(9/25)*x²ⁿ⁺²*y⁴-(26/36)*x⁴ⁿ⁻²*y²ⁿ⁺²+x³ⁿ*yⁿ⁺².
(9/25)*x²ⁿ⁺²*y⁴≡(9/25)*x²ⁿ⁺²*y⁴.
2x - <span> </span><span>песок</span>
<span>5x - </span><span>чернозеm</span>
<span>2x + 5x = 42</span>
<span>7x = 42</span>
<span>x = 6</span>
<span>
</span>
<span>2 * 6 = 12 - pesok</span>
<span>5 * 6 = 30 - chernozem</span>
<span>sin5π/12×cosπ/12+sinπ/12×cos5π/12=sin(5</span>π/12+π/12)=sin6π/12=sinπ/2=1.
Используем формулу: 1+tg^2 a=1/cos^2 a, ответ : -sqrt5/2