Выражение: (0.1*x^3-0.3*y)*(0.1*x^3-0.3*y)
Ответ: 0.01*x^6-0.06*x^3*y+0.09*y^2
Решаем по действиям:
1. (0.1*x^3-0.3*y)*(0.1*x^3-0.3*y)=0.01*x^6-0.06*x^3*y+0.09*y^2
(0.1*x^3-0.3*y)*(0.1*x^3-0.3*y)=0.1*x^3*0.1*x^3-0.1*x^3*0.3*y-0.3*y*0.1*x^3+0.3*y*0.3*y
1.1. 0.1*0.1=0.01
X0.1
_0_._1_ _
01
0_0_ _ _
0.01
1.2. x^3*x^3=x^6
x^3*x^3=x^(3+3)
1.2.1. 3+3=6
+3
_3_
6
1.3. 0.1*0.3=0.03
X0.1
_0_._3_ _
03
0_0_ _ _
0.03
1.4. 0.3*0.1=0.03
X0.3
_0_._1_ _
03
0_0_ _ _
0.03
1.5. -0.03*x^3*y-0.03*y*x^3=-0.06*x^3*y
1.6. 0.3*0.3=0.09
X0.3
_0_._3_ _
09
0_0_ _ _
0.09
1.7. y*y=y^2
y*y=y^(1+1)
1.7.1. 1+1=2
+1
_1_
2
Решаем по шагам:
1. 0.01*x^6-0.06*x^3*y+0.09*y^2
1.1. (0.1*x^3-0.3*y)*(0.1*x^3-0.3*y)=0.01*x^6-0.06*x^3*y+0.09*y^2
(0.1*x^3-0.3*y)*(0.1*x^3-0.3*y)=0.1*x^3*0.1*x^3-0.1*x^3*0.3*y-0.3*y*0.1*x^3+0.3*y*0.3*y
1.1.1. 0.1*0.1=0.01
X0.1
_0_._1_ _
01
0_0_ _ _
0.01
1.1.2. x^3*x^3=x^6
x^3*x^3=x^(3+3)
1.1.2.1. 3+3=6
+3
_3_
6
1.1.3. 0.1*0.3=0.03
X0.1
_0_._3_ _
03
0_0_ _ _
0.03
1.1.4. 0.3*0.1=0.03
X0.3
_0_._1_ _
03
0_0_ _ _
0.03
1.1.5. -0.03*x^3*y-0.03*y*x^3=-0.06*x^3*y
1.1.6. 0.3*0.3=0.09
X0.3
_0_._3_ _
09
0_0_ _ _
0.09
1.1.7. y*y=y^2
y*y=y^(1+1)
1.1.7.1. 1+1=2
+1
_1_
2
Пусть x - первое число, y - второе число. Составляем систему уравнений согласно условию:
![\left \{ {{x^2 - y^2=100} \atop {3x -2y=30}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bx%5E2+-+y%5E2%3D100%7D+%5Catop+%7B3x+-2y%3D30%7D%7D+%5Cright.+)
Из второго уравнения выразим икс и подставим его в первое уравнение:
![x=10+ \frac{2}{3} y \\ \\ (10+ \frac{2}{3} y)^2 -y^2 =100 \\ \\ 100 +\frac{40}{3} y + \frac{4}{9} y^2 -y^2 = 100 \\ \\ \frac{40}{3} y -\frac{5}{9} y^2 =0 \\ \\ 24y -y^2 =0 \\ \\ y(24-y)=0 \\ \\ y_1=0 \\ y_2=24](https://tex.z-dn.net/?f=x%3D10%2B+%5Cfrac%7B2%7D%7B3%7D+y+%5C%5C++%5C%5C+%2810%2B+%5Cfrac%7B2%7D%7B3%7D+y%29%5E2+-y%5E2+%3D100+%5C%5C++%5C%5C+100+%2B%5Cfrac%7B40%7D%7B3%7D+y+%2B+%5Cfrac%7B4%7D%7B9%7D+y%5E2+-y%5E2+%3D+100+%5C%5C++%5C%5C+%5Cfrac%7B40%7D%7B3%7D+y+-%5Cfrac%7B5%7D%7B9%7D+y%5E2+%3D0+%5C%5C++%5C%5C+24y+-y%5E2+%3D0+%5C%5C++%5C%5C+y%2824-y%29%3D0+%5C%5C++%5C%5C+y_1%3D0+%5C%5C+y_2%3D24)
Определяем икс:
![x_1 =10+ \frac{2}{3} y_1 = 10+ \frac{2}{3}*0 =10 \\ \\ x_2 =10+ \frac{2}{3} y_2 = 10+ \frac{2}{3}*24 =26](https://tex.z-dn.net/?f=x_1+%3D10%2B+%5Cfrac%7B2%7D%7B3%7D+y_1+%3D+10%2B+%5Cfrac%7B2%7D%7B3%7D%2A0+%3D10+%5C%5C++%5C%5C+x_2+%3D10%2B+%5Cfrac%7B2%7D%7B3%7D+y_2+%3D+10%2B+%5Cfrac%7B2%7D%7B3%7D%2A24+%3D26)
Итак:
![x_1 =10; y_1=0 \\ \\ x_2=26; y_2= 24](https://tex.z-dn.net/?f=x_1+%3D10%3B+y_1%3D0+%5C%5C++%5C%5C+x_2%3D26%3B+y_2%3D+24)
Проверка показывает, что оба решения верны.
(x-1)^2*(x^2+4x-12)<0
(x-1)^2*(x^2+4x-12)=0
(x-1)^2=0 или x^2+4x-12=0
для 1: x(1)=1
для 2: x(2)=2
x(3)=-6
Подстановка:
-7: (-7-1)^2*(-7^2-7*4-12)=64*12>0
0: (0-1)^2*(-12)<0
1,5: (1,5-1)^2*(1,5^2+1,5*4-12)<0
3: (3-1)^2*(3^2+3*4-12)>0
Ответ: -6<x<1 и 1<x<2
№3. а)
![x^{2} -8x+12 = x^{2} -8x+16-4=(x-4)^{2} -2^{2}](https://tex.z-dn.net/?f=+x%5E%7B2%7D+-8x%2B12+%3D+x%5E%7B2%7D+-8x%2B16-4%3D%28x-4%29%5E%7B2%7D+-2%5E%7B2%7D+)
= (x-4-2)(x-4+2) = (x-6)(x-2)
б)
![9 x^{2} -6x-8 = 9 x^{2} -6x+1-9 = (3x-1)^{2} -3^{2}](https://tex.z-dn.net/?f=9+x%5E%7B2%7D+-6x-8+%3D+9+x%5E%7B2%7D+-6x%2B1-9+%3D+%283x-1%29%5E%7B2%7D+-3%5E%7B2%7D)
= (3x-1-3)(3x-1+3) = (3x-4)(3x+2)
№ 2. а) x²-6x+11 = x²-6x+9+2 = (x-3)²+2
б) x²+6x = x²+6x+9-9= (x+3)²-9