1 [1]. Solve the following system of equations $$ \left\{ \begin{array}{llllll} x_1&+\lambda_1x_2&+\lambda_1^2x_3&+\lambda_1^4x_4&+\lambda_1^6x_5&=0 \\ x_1&+\lambda_2x_2&+\lambda_2^2x_3&+\lambda_2^4x_4&+\lambda_2^6x_5&=0 \\ x_1&+\lambda_3x_2&+\lambda_3^2x_3&+\lambda_3^4x_4&+\lambda_3^6x_5&=0 \\ x_1&+\lambda_4x_2&+\lambda_4^2x_3&+\lambda_4^4x_4&+\lambda_4^6x_5&=0 \\ x_1&+\mu x_2&+\mu^2x_3&+\mu^4x_4&+\mu^6x_5&=1\, . \end{array} \right. $$

2 [2]. The following equality is valid: $$ \frac{1}{(1-x_1x)\times \dots \times (1-x_mx)}=1+\sigma_{1} (x_1,\dots,x_m)x+\sigma_{2} (x_1,\dots,x_m)x^2+\dots $$

3 [2]. Function $ \sigma_{2k} (x_1,\dots,x_m), k \in \mathbb N $, treated as a homogeneous polynomial in its variables, is positively definite.

4 [2]. Let $ \alpha_1< \alpha_2<\dots< \alpha_n, \beta_1< \beta_2<\dots< \beta_n $. Then $$ \left| \begin{array}{cccc} e^{\alpha_1 \beta_1} & e^{\alpha_1 \beta_2} & \dots & e^{\alpha_1 \beta_n} \\ e^{\alpha_2 \beta_1} & e^{\alpha_2 \beta_2} & \dots & e^{\alpha_2 \beta_n} \\ \vdots & & & \vdots \\ e^{\alpha_n \beta_1} & e^{\alpha_n \beta_2} & \dots & e^{\alpha_n \beta_n} \end{array} \right|>0 \, . $$

5 . Prove that the determinants $$ \left| \begin{array}{ccc} 1 & x_1 & x_2x_3 \\ 1 & x_2 & x_1x_3 \\ 1 & x_3 & x_1x_2 \end{array} \right|, \ \left| \begin{array}{cccc} 1 & x_1 & x_1^2 & x_2x_3x_4 \\ 1 & x_2 & x_2^2 & x_1x_3x_4 \\ 1 & x_3 & x_3^2 & x_1x_2x_4 \\ 1 & x_4 & x_4^2 & x_1x_2x_3 \end{array} \right|, \ \left| \begin{array}{ccccc} 1 & x_1 & x_1^2 & x_1^3 & x_2x_3x_4x_5 \\ 1 & x_2 & x_2^2 & x_2^3 & x_1x_3x_4x_5 \\ 1 & x_3 & x_3^2 & x_3^3 & x_1x_2x_4x_5 \\ 1 & x_4 & x_4^2 & x_4^3 & x_1x_2x_3x_5 \\ 1 & x_5 & x_5^2 & x_5^3 & x_1x_2x_3x_4 \end{array} \right|, \dots $$ coincide, up to a sign, with Vandermonde's determinants of the corresponding orders.

[1]. Лефорт Г. Алгебра и анализ. Задачи. М.Наука. 1973, c.89

[2]. Pólya G. , Szegö G.Problems and Theorems in Analysis. V.2. Springer. 1998