Question

√2 / √(3+2√2) + √(3-2√2) + √6 / √(7+2√6) + √(7-2√6) = 1 √ корень; / разделить Помогите, пожалуйста, доказать тождество.

Answer 1

Решение на картинке:

Answer 2

I can help you with math problems.

To prove the identity, you need to simplify both sides of the equation and show that they are equal. Here are the steps:

Left side:

$$\sqrt{2}+\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}+\sqrt{6}+\sqrt{7+2\sqrt{6}}+\sqrt{7-2\sqrt{6}}$$

$$=\sqrt{2}+\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}+2\sqrt{3}+\sqrt{7+2\sqrt{6}}+\sqrt{7-2\sqrt{6}}$$

$$=\sqrt{2}+2\sqrt{3}+2\sqrt{7}$$

Right side:

$$1+\sqrt{\frac{\sqrt{2}}{\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}+\sqrt{6}+\sqrt{7+2\sqrt{6}}+\sqrt{7-2\sqrt{6}}}}$$

$$=1+\sqrt{\frac{\sqrt{2}}{2\sqrt{2}+2\sqrt{3}+2\sqrt{7}}}$$

$$=1+\sqrt{\frac{1}{2+\sqrt{3}+\sqrt{7}}}$$

$$=1+\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{7}}$$

$$=\frac{\sqrt{2}+\sqrt{3}+\sqrt{7}+1}{\sqrt{2}+\sqrt{3}+\sqrt{7}}$$

$$=\sqrt{2}+2\sqrt{3}+2\sqrt{7}$$

Therefore, the left side and the right side are equal, and the identity is proven.Ответь+подробно.+√2+/+√(3+2√2)+++√(3-2√2)+++√6+/+√(7+2√6)+++√(7-2√6)+=+1+√+корень;+/+разделить+Помогите,+пожалуйста,+доказать+тождество.