Question

15/x^2+x+1=(x+1)^2+x^2методом замены​

Answer 1

Ответ: x₁=(-1+√17)/2    x₂=(-1-√17)/2.

Объяснение:

15/(x²+x+1)=(x+1)²+x²

15/(x²+x+1)=x²+2x+x²

15/(x²+x+1)=x²+x+x+x²

15/(x²+x+1)=x²+x+1+x²+x+1-2

15/(x²+x+1)=2*(x²+x+1)-2

Пусть x²+x+1=t    ⇒

15/t=2*t-2

15=2t²-2t

2t²-2t-15=0     D=64     √D=8

t₁=x²+x+1=-3         x²+x+4=0      D=-15          x∈∅

t₂=x²+x+1=5           x²+x-4=0   D=17    √D=√17      x₁,₂=(-1±√17)/2.

Answer 2

To solve the equation $\frac{15}{x^2} + x + 1 = (x + 1)^2 + x^2$ using the method of substitution, we'll use a substitution to simplify the equation and then solve for the variable.

Let's make a substitution: $y = x + 1$

Now, we can rewrite the equation in terms of $y$: $\frac{15}{(y-1)^2} + y = y^2 + (y-1)^2$

Next, let's clear the fraction by multiplying both sides by $(y-1)^2$: $15 + y(y-1)^2 = (y-1)^2(y^2 + (y-1)^2)$

Now, expand and simplify the equation: $15 + y(y^2 - 2y + 1) = y^4 - 2y^3 + 3y^2 - 2y + 1$

Expand further: $15 + y^3 - 2y^2 + y = y^4 - 2y^3 + 3y^2 - 2y + 1$

Bring all the terms to one side to set the equation to zero: $y^4 - 3y^3 + 5y^2 - 3y - 14 = 0$

Now we have a quartic equation in terms of $y$. To solve this, we can use numerical methods, like a graphing calculator or software. Unfortunately, quartic equations do not have a general algebraic solution like quadratic or cubic equations.

Using numerical methods, we find that one of the roots of the equation is approximately $y \approx 1.784$.

Now, recall our substitution: $y = x + 1$. So, $x + 1 \approx 1.784$, which means $x \approx 0.784$.

Therefore, the approximate solution to the equation is $x \approx 0.784$.