Question

(4√2-√18+√128):√162=

Answer 1

Ответ:

1

решение смотрите на фотках

Answer 2

To simplify the expression $\frac{{4\sqrt{2} - \sqrt{18} + \sqrt{128}}}{{\sqrt{162}}}$, we'll start by simplifying the radicals in the numerator and denominator.

1. Simplify the radicals in the numerator: $\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$ $\sqrt{128} = \sqrt{64 \times 2} = 8\sqrt{2}$

   So, the numerator becomes $4\sqrt{2} - 3\sqrt{2} + 8\sqrt{2} = 9\sqrt{2}$.

2. Simplify the radical in the denominator: $\sqrt{162} = \sqrt{81 \times 2} = 9\sqrt{2}$.

Now, we can rewrite the expression with the simplified radicals:

$$\frac{{9\sqrt{2}}}{{9\sqrt{2}}} = 1.$$

So, $\frac{{4\sqrt{2} - \sqrt{18} + \sqrt{128}}}{{\sqrt{162}}} = 1$.