Ответ:
fucjckvkvbbckvkvkvckckvkvjvkv
равняется среднему геометрическому
и
Ответ: 12
1) (^x-^y)(x+^x^y+y)/(^x-^y)=x+^x^y+y
2) (^x+^y)/(^x+^y)(x-^x^y+y)=1/(x-^x^y+y)
3) <span>(2^2-x^x) / (2+^(2x)+x )=(^2-^x)((2+^(2x)+x ))/((2+^(2x)+x ))=(^2-^x)</span>
<span>4) (<span>a-^(3a) +3) / a^a+3^3=(<span>a-^(3a) +3) / )(^a-^3)(a-^(3a) +3)</span></span></span>
<span><span><span>5) <span>(2^10-5) / 4-^10 = ^5*(2^2-^5)/^2*(2^2-^5)=^5/^2</span></span></span></span>
<span><span><span><span>6) (<span>9-2^3) / (3^6-2^2) = ^3*(3^3-2)/(3^3-2)*^2=^3/^2</span></span></span></span></span>
<span><span><span><span><span>7) (<span>^70-^30) / ^35-^15 = ^2(^35-^15)/^35-^15=^2</span></span></span></span></span></span>
<span><span><span><span><span><span>8) <span>^15-5 / ^6-^10= ^5(^3-^5)/^2(^3-^5)</span></span></span></span></span></span></span>
<span><span><span><span><span><span><span>9) <span> (^10-1) в квадрате 3 / ^10+^3-1 - не указали каким боком тройка сюда относится</span></span></span></span></span></span></span></span>
A2/a1=a7/a2 => a1*(a1+6d)=(a1+d)^2 => d=4a1 =>a5=a1+4*d=17a1 ответ:17