Question

√(6 + 2√5) - √(6 - 2√5) = ?

Answer 1

$\sqrt{6+2 \sqrt{5} }- \sqrt{6-2 \sqrt{5} } = \sqrt{5+2 \sqrt{5}+1 } - \sqrt{5-2 \sqrt{5}+1 }=$ $\sqrt{( \sqrt{5}+1) ^{2} } - \sqrt{( \sqrt{5}-1) ^{2} }=| \sqrt{5}+1|-| \sqrt{5} -1|= \sqrt{5} +1- \sqrt{5}+1=$ $2$

Answer 2

Добавлю второй способ решения этой задачи

Answer 3

Let's simplify this expression step by step:

√(6 + 2√5) - √(6 - 2√5)

= √(23 + 2√5) - √(23 - 2√5) (replacing 6 with 2*3)

= √2√(3 + √5) - √2√(3 - √5) (taking out a common factor of √2)

= √2* (√(3 + √5) - √(3 - √5)) (factoring out √2)

Now, we need to simplify the expression inside the square root. Let's call it A:

A = √(3 + √5) - √(3 - √5)

We can rationalize the denominator of each square root expression by multiplying the numerator and denominator by the conjugate of the denominator:

A = (√(3 + √5) - √(3 - √5)) * (√(3 + √5) + √(3 - √5)) / (√(3 + √5) + √(3 - √5))

Simplifying the numerator using the difference of squares:

A = (3 + √5 - (3 - √5)) / (√(3 + √5) + √(3 - √5))

A = (2√5) / (√(3 + √5) + √(3 - √5))

Now we can substitute this value back into our original expression:

√2* (√(3 + √5) - √(3 - √5))

= √2 * A

= √2 * [(2√5) / (√(3 + √5) + √(3 - √5))]

Simplifying the expression by rationalizing the denominator:

= √10 * [(√(3 + √5) - √(3 - √5)) / (2)]

= √10 * A / 2

= √10 * (2√5) / [2 * (√(3 + √5) + √(3 - √5))]

Simplifying again by cancelling the common factor of 2:

= √5 / (√(3 + √5) + √(3 - √5))

Therefore, the value of the expression is:

√(6 + 2√5) - √(6 - 2√5) = √5 / (√(3 + √5) + √(3 - √5))