В) x^2+2x+x+2=2x^2-4x-x+2
-x^2+8x=0
x(-x+8)=0
x1=0 x2=8
А) = 7(a + b)
б) = 8(-b +c)
в) = 12(x +4y)
г) = -9(m +3n)
д) = 12(a + 1)
e) = -10(1 +c)
<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>
Log ( 1/2 ) ( 3x - 5 ) = 2
1/2 ^ 2 = 3x - 5
0,25 = 3x - 5
3x = 5,25
X = 1,75