Current location - Training Enrollment Network - Mathematics courses - The eighth part of junior high school mathematics problems
The eighth part of junior high school mathematics problems
Solution: 1 6√ 15/(3√2+2√3)

= 6√ 15(3√2-2√3)/(3√2+2√3)(3√2-2√3)

=( 18√30-36√5)/( 18- 12)

=( 18√30-36√5)/6

=3√30-6√5

2.(5√3-3√6)/(3√3+2√6)

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

=(45-30√2-27√2+36)/(27-24)

=(8 1-57√2)/3

=27- 19√2

3.( 1/√3+√2)+( 1/√2- 1)-2/√3+ 1

= √3/3+√2+√2/2- 1-2√3/3+ 1

=3√2/2-√3/3

4.( 1/(a√(9a? ))-√b? /(3b? ))-(2a√( 1/4a) -b√(25/b? I don't know which ones are on the numerator and which ones are on the denominator.

=√a/3a? -√b/3b-√a-5√b/b

=√a( 1- 1/3a? )-√b( 1/3b- 15/3b)

=√a( 1- 1/3a? )- 16√b/3b

5.x? -4x- 1=(x+2)? -8x-5

=( 1/√5)? -8( 1/√5-2)-5

= 1/5-8√5/5+ 16-5

=56/5-8√5/5