2sin5xcos2x-2cos2x=0
2cos2x(sin5x-1)=0
cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2
<span>sin5x=1⇒5x=π/2+2πn⇒x=π/10+πn/5
</span>Проверь я могу ошибаться
1) 3(b-1)² +8b =3(b² +2b+1)+8b= 3b² +6b+3+8b= 3b² +14b+3;
2) (n+5)² -n(n-7)= n² +10n +25 -n² +7n= 17n+25;
3) 2c(8c-3)-(4c+1)² = 16c-6c-(16c²+8c+1)= 10c-16c²-8c-1= 2c-16c²-1;
4) (6-n)(n+6)+(n-4)² = n² -36+n² +8n+16= 2n² +8n-20;
(√5-1)^2(2√5+1)^2=5-2√5+1-20-4√5-1=-6√5-15
1. <span>2xy² - 18х = </span> 2х(<span>y² - 9) = </span> 2х(y - 3)(y + 3)
2. <span>(b-1)²(b+2) - b²(b-3) + 3</span> = (<span>b² - 2</span><span>b + 1)</span>(b+2) - <span>b³ + 3</span>b² + 3 =
= <u /><span><u>b³ +</u> 2</span>b² - 2<span>b² - 4</span>b + <span>b + 2<u /></span><u> - b³</u> + 3b² + 3 =
= - 4b + <span>b + 2</span> + 3b² + 3 = 3b² + 5<span>b + 5</span>