=-4(3√2)(2√3)

=-24√6

2.√3÷√2×( 14/(3-√2)-(√24+√ 12)

=√(3/2)* 14(3+√2)/[(3-√2)(3+√2)]-2√6-2√3

=√6/2*(6+2√2)-2√6-2√3

=3√6+2√3-2√6-2√3

=√6

3. Simplification: [(a-2√ab+b)/√a-√b]-(√a+√b)

=(√a-√b)^2/(√a-√b)-(√a+√b)

=(√a-√b)-(√a+√b)

=-2√b

4. Given A-B = 2+√ 3 and B-C = 2-√ 3, find the algebraic expression.

The value of a 2+b 2+c 2-ab-BC-ca.

a^2+b^2+c^2-ab-bc-ca

= 1/2[(a-b)^2+(b-c)^2+(c-a)^2]

= 1/2[(a-b)^2+(b-c)^2+(a-b+b-c))^2]

= 1/2*[(2+√3)^2+(2-√3)^2+(2+√3+2-√3)^2]

= 1/2*(8+6+ 16)

= 15

5. let n be a natural number and x >；; 1, compare the sizes of (√n+ 1)-√n and √n-(√n- 1).

(√n+ 1)-√n = 1/[√( n+ 1)+√n]

√n-(√n- 1)= 1/[√n+√( n- 1)]

Because [√ (n+1)+√ n] > [√n+√(n- 1)]

So 1/[√ (n+ 1)+√ n]

(√n+ 1)-√n & lt； √n-(√n- 1)