Question

Помогите пожалуйста якласс 10 класс 6 неделя

Answer 1

изи) удачи )))).....

Answer 2

$\displaystyle\tt\left(\frac{x^{0.5}-3}{x^{0.5}+3}+ \frac{x^{0.5}+3}{x^{0.5}-3}-\frac{36}{x-9}\right)^3= \left(\frac{(x^{0.5}-3)^2+(x^{0.5}+3)^2}{(x^{0.5}+3)(x^{0.5}-3)}-\frac{36}{x-9}\right)^3= \\\\\\=\left(\frac{x-6x^{0.5}+9+x+6x^{0.5}+9}{(x^{0.5})^2-(3)^2}-\frac{36}{x-9}\right)^3=\left(\frac{2x+18}{x-9}-\frac{36}{x-9}\right)^3=\\\\\\=\left(\frac{2x+18-36}{x-9}\right)^3=\left(\frac{2x-18}{x-9}\right)^3=\left(\frac{2(x-9)}{x-9}\right)^3=2^3=8$