<span><span>x4</span> + 3<span>x2</span> - 10 = 0</span>
<span>Сделаем замену </span>y<span> = </span>x2, тогда биквадратное уравнение примет вид
<span><span>y2</span> + 3y - 10 = 0</span>
Для решения этого квадратного уравнения найдем дискриминант:
D = 32<span> - 4·1·(-10) = 49</span>
<span><span><span>y1 = </span><span>-3 - √49</span> = -5</span>2·1</span><span><span><span>y2 = </span><span>-3 + √49</span> = 2</span>2·1</span>
<span><span>x2 = </span>-5</span><span><span>x2 = </span>2</span>
<span><span>x1</span> = <span>√2</span></span><span><span>x2</span> = -<span>√<span>2</span></span></span>
(2х-12)4=3х+2;
8х-48-3х-2=0;
5х=50;
х=10.
1) 64а^2-9х^2+6х-1=(64а^2-1)-3х(3х-2)=(8а-1)(8а+1)-3х(3х-2);;;;;;;2) 25-m^2-8mn-16n^2=(25-16n^2)-m(m+8n)=(5-4n)(5+4n)-m(m+8n);;;;3) a^6+1+3а^2+3а^4=(а^3+1)^2+3а^2(1+а^2)