X*(x+1)*(x+2)*(x+3) = (x*(x+3))*((x+1)*(x+2)) = ( x^2 +3x)*( x^2 +3x+2) = = 120,
сделаем замену x^2+3x = y, тогда
y*(y+2) =120;
y^2 + 2y - 120 = 0;
D/4 = 1+120 = 121=11^2;
y1 = (-1+11) = 10;
y2 = (-1-11) = -12;
1) x^2 + 3x = 10;
x^2 + 3x - 10 = 0;
D = 9 - 4*(-10) = 49 = 7^2;
x1 = (-3+7)/2 = 4/2 = 2;
x2 = (-3-7)/2 = -10/2 = -5;
2) x^2 + 3x = -12;
x^2 + 3x + 12 = 0;
D = 9 - 4*12 < 0; в 2) решений нет.
Ответ. x1 = 2; x2 = -5;
(sina)^2+(cosa)^2=1 >>> (cosa)^2=1-8/9=1/9>>>>cosa=1/3 6*cosa=2
Ответ:
∆=5*12-3*а=0, 60=3а, а=20 -ответ
4-5(6x-6)<54-40x
4-30x+30<54-40x
-30x+40x<54-4-30
10x<20
x<20/10=2
x<2
2x²+4y²=24
4x²+8y²=24x⇒2x²+4y²=12x
12x=24
x=2
2*4+4y²=24
4y²=16
y²=4
y=-2 U y=2
(2;-2) U (2;2)