1)2m(3y-1)+3(3y-1)=(3y-1)(2m+3)
2)x(a-3)-4(a-3)=(a-3)(x-4)
4)9y^2+30y+25=3y(3y+10)+5(6y+5)
<span>Sin^2(3pi/2-y)+sin^2(3pi+y)+2tg(5pi/2-y)×tg (3pi+y)
</span>---------------------------------
Решение
Вычислить Sin²(3π/2-y)+sin²(3π+y)+2tg(5π/2-y)*tg (3π+y) .
<span>---</span>
Применяя формулы приведения ,получаем :
Sin²(3π/2-y)+sin²(3π+y)+2tg(5π/2-y)*tg (3π+y)= (cos²y +sin²y)+ 2ctqy*tqy =1 +2*`1 =3.
ответ : 3.
===========================
sin (3π<span>/2-y)) = -cosy
</span>sin(3π +y ) =sin( (2π+(π +y ) )=sin(π<span> +y </span><span>) = </span> -siny
tg(5π/2-y) =tg((2π +(π<span>/2-y) )</span> =tq(π/2-y) =ctqy
tq(3π +y ) =tq( (2π+(π +y ) ) =tq(π<span> +y ) =tqy</span>
= tg 30 = (корень из 3)/3
(a+1)³-8a⁶= (a+1-2a²)*(a²+2a+1+(a+1)*2a²+(2a²)²)= (a+1-2a²)(a²+2a+1+2a³+2a²+4a⁴)= (a+1-2a²)(4a⁴+2a³+3a²+2a+1) или если (a+1-2a²) разложить еще на множители, то получится:
(1-a)(2a+1)(4a⁴+2a³+3a²+2a+1)