Question

Помогите решить 1) (6^1/2+2^1/2)^2+(6^1/2-2^1/2)^2 2) x-y/x^1/2+y^1/2

Answer 1

6+2×12^1/2+2+6-2×12^1/2+2=6+2+6+2=16

Answer 2

1. Раскроем скобки с помощью формулы $(a+b)^2 = a^2 + 2ab + b^2$: \begin{align*} &(6^{1/2}+2^{1/2})^2+(6^{1/2}-2^{1/2})^2 \ &= 6 + 2 \cdot 6^{1/2} \cdot 2^{1/2} + 2 + 6 - 2 \cdot 6^{1/2} \cdot 2^{1/2} + 2 \ &= 6 + 4 + 2 + 6 - 4 - 2 + 2 \ &= 18. \end{align*} Таким образом, $(6^{1/2}+2^{1/2})^2+(6^{1/2}-2^{1/2})^2 = 18$.

2. Для удобства воспользуемся свойством $(a-b)(a+b) = a^2 - b^2$. Тогда: \begin{align*} \frac{x-y}{x^{1/2}+y^{1/2}} &= \frac{(x-y)(x^{1/2}-y^{1/2})}{(x^{1/2}+y^{1/2})(x^{1/2}-y^{1/2})} \ &= \frac{x^{3/2}-xy^{1/2}-yx^{1/2}+y^{3/2}}{x-y} \ &= \frac{(x^{3/2}+y^{3/2})-xy(x^{-1/2}+y^{-1/2})}{x-y}. \end{align*} Таким образом, $\frac{x-y}{x^{1/2}+y^{1/2}} = \frac{(x^{3/2}+y^{3/2})-xy(x^{-1/2}+y^{-1/2})}{x-y}$.