Faster (and probably easier): Note that the denominator can be written as: 5 - √(4 * 6) = 5 - 2√6 = 5 - 2√3√2 = 3 + 2 - √3√2 = (√3)² + (√2)² - 2√3√2 = (√3-√2)² The numerator is obviously 5² So the whole root is a ratio of two squares. Eliminating root and squares, 5/(√3-√2) rationalising the denominator by multiplying by (√3+√2)/(√3+√2) gives 5(√3+√2)/[((√3)²-(√2)²] = 5(√3+√2) / 1 = 5(√3+√2)
5-sqrt(24)=5-2sqrt(6)
=[sqrt(3)-sqrt(2))²
Thus sqrt[25/{5-sqrt(24)}]
=5/[sqrr(3)-sqrt(2)]
=5[sqrt(3)+sqrt(2)]
√(125+25√24) =√(25（5+√24) ）=5√(5+√24) =5√(5+√(4×6)) =5√(5+2√(3×2))
=5√(2+3+2√(2×3)) =5（√2+√3）
Faster (and probably easier):
Note that the denominator can be written as:
5 - √(4 * 6) = 5 - 2√6 = 5 - 2√3√2 = 3 + 2 - √3√2 = (√3)² + (√2)² - 2√3√2 = (√3-√2)²
The numerator is obviously 5²
So the whole root is a ratio of two squares. Eliminating root and squares,
5/(√3-√2)
rationalising the denominator by multiplying by (√3+√2)/(√3+√2) gives
5(√3+√2)/[((√3)²-(√2)²] = 5(√3+√2) / 1 = 5(√3+√2)
Спасибо за учебу!
Unbelievably long-winded way to do this
I want to add this point
a+√(a^2-1)=[√{(a+1)/2}+√{(a^2-1)/2}]^2 where a is odd.
Take 7+√48=(√4=√3)^2=(2+√3)^2
This is solved in the mind and is approximately 15.73
√25=5,√(5_√24)=√(√3-√2)^2=√3-√2
So answer is 5/(√3-√2)=5(√3+√2)/(3-2)=5(√3+√2).
Or take answer as x. First find x^2,then x.
2중근호만 알면 암산으로 5초
=√(125+25√24/25+5√24-5√24-24)
=√(125+25√24/(1))
(100+25)+(25.2√6)
=√[25(5)+2√6)]
This Indian is too long-winded. He just took out the 25 in the square root.
What a drag!