Question

1) √6+√3 дробь√3 -√2 2)√2-1дробь√2+1 +√2 3)√4-√2 дробь 2 +3√2 дробь6 4) 3+2√2 дробь√2+1 -√2 с решением пожалуйста какие рациональные какие иррациональные?

Answer 1

Вот и всё, всё решил
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Answer 2

We can simplify the expression as follows:

√6 + √3 = (√3 * √2) + √3 = √3(√2 + 1)

Thus, the expression becomes:

(√3(√2 + 1))/(√3 * √3 - √2) = (√3(√2 + 1))/(3 - √2)

This is a rationalized denominator, since the denominator contains only rational numbers (3 and √2), and the numerator contains the irrational number √3.

We can simplify the expression as follows:

(√2 - 1)/(√2 + 1) + √2 = ((√2 - 1) + √2(√2 + 1))/(√2 + 1) = (√2 - 1 + 2√2)/(√2 + 1)

To rationalize the denominator, we can multiply both the numerator and denominator by (√2 - 1):

(√2 - 1 + 2√2)/(√2 + 1) * (√2 - 1)/(√2 - 1) = (2√2 + 1)/(1) = 2√2 + 1

This expression is irrational, since it contains the irrational number √2.

We can simplify the expression as follows:

(√4 - √2)/2 + (3√2)/6 = (√4 - √2)/2 + √2/2 = (√4 + √2)/2

This expression is irrational, since it contains the irrational number √2.

We can simplify the expression as follows:

(3 + 2√2)/(√2 + 1) - √2 = (3 + 2√2)/(√2 + 1) - (√2(√2 + 1))/(√2 + 1) = (3 + 2√2 - 2 - 2√2)/(√2 + 1) = 1/(√2 + 1)

To rationalize the denominator, we can multiply both the numerator and denominator by (√2 - 1):

1/(√2 + 1) * (√2 - 1)/(√2 - 1) = (√2 - 1)/(2 - 1) = √2 - 1

This expression is irrational, since it contains the irrational number √2.