Question

6(x²+1/x²)+5(x+1/x)-22=0​

Answer 1

Ответ:

Пошаговое объяснение

      -17- корень 145               -17 + корень 145

x1=------------------------------ , x2= ---------------------------------- , x3= 1

                12                                         12

x1~ - 2,42013, x2~ -0,4132, x3 = 1

Answer 2

To solve the equation 6(x² + 1/x²) + 5(x + 1/x) - 22 = 0, we can simplify it by first eliminating the fractions. Multiply the entire equation by x² to get rid of the denominators:

6(x²)(x² + 1/x²) + 5(x²)(x + 1/x) - 22(x²) = 0

Simplifying further:

6(x⁴ + 1) + 5(x³ + 1) - 22x² = 0

Now, distribute the coefficients:

6x⁴ + 6 + 5x³ + 5 - 22x² = 0

Combine like terms:

6x⁴ + 5x³ - 22x² + 11 = 0

This equation is a quartic equation (a polynomial of degree 4). Solving quartic equations generally involves more advanced methods, such as factoring, using the rational root theorem, or employing numerical methods. In some cases, quartic equations may not have exact algebraic solutions.

If you have any specific values or constraints for x, please let me know, and I can assist you further.