Решите пожалуйста уравнения: (3x+1)^2=5(x+1/3)

(x+1/2)^2=2x+1

(x-1/7)^2=3(7x-1)

x+x^2=x^3+x^4

x-x^2=x^3-x^4

x^3+x^2+x+1=0

x^3-x^2+x-1=0

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Ответы и объяснения

2012-12-26T16:29:28+00:00

Ответ: 9*x^2+x-(2//3)=0Выражение: (3*x+1)^2=5*(x+1/3)

Ответ: 9*x^2+x-(2//3)=0

Решаем по действиям:
1. (3*x+1)^2=9*x^2+6*x+1
(3*x+1)^2=((3*x+1)*(3*x+1))
1.1. (3*x+1)*(3*x+1)=9*x^2+6*x+1
(3*x+1)*(3*x+1)=3*x*3*x+3*x*1+1*3*x+1*1
1.1.1. 3*3=9
X3
_3_
9
1.1.2. x*x=x^2
x*x=x^(1+1)
1.1.2.1. 1+1=2
+1
_1_
2
1.1.3. 3*x+3*x=6*x
2. 1/3=(1//3)
3. 5*(x+(1//3))=5*x+(5//3)
5*(x+(1//3))=5*x+5*(1//3)
3.1. 5*(1//3)=(5//3)
4. 9*x^2+6*x+1-(5*x+(5//3))=9*x^2+6*x+1-5*x-(5//3)
5. 6*x-5*x=1*x
6. 1-(5//3)=-(2//3)