F.2 Maths

a^2-25x^2+10x-1
= a^2 - (25x^2 - 10x + 1)
= a^2 - [(5x)^2 - 2(5)(1)x + 1^2]
= a^2 - (5x-1)^2
= (a-5x+1)(a+5x-1)
1a. Factorize 98(a-b)^2-18(a+b)^2
= 2[49(a-b)^2 - 9(a+b)^2]
= 2[7(a-b) - 3(a+b)] [7(a-b) + 3(a+b)]
= 2(4a - 10b)(10a - 4b)
= 2 * 2(2a - 5b) * 2(5a - 2b)
= 8(2a - 5b)(5a - 2b)
bi)Factorize 25x^2-20x+4
= (5x)^2 - 2(5x)(2) + 2^2
= (5x - 2)^2
bii)Hence,factorize 25(m-1)^2-20m+24
= 25(m-1)^2 - 20m + 20 + 4
= 25(m-1)^2 - 20(m-1) + 4
Compare with i): i.e. let m-1 = x in i) :
25(m-1)^2 - 20(m-1) + 4
= [5(m-1) - 2]^2
= (5m - 7)^2
c)I think your queation should be :
Prove that the multiple of four consecutive integers plus 1 is a perfect square.

Let four consecutive integers be x-1 ,x ,x+1 ,x+2 ,then
(x-1)x(x+1)(x+2) + 1
= x(x+2)(x^2 - 1) + 1
= x(x^3 + 2x^2 - x - 2) + 1
= x^4 + 2x^3 - x^2 - 2x + 1
= (x^2 + x - 1)^2 is a perfect square.

a^2-25x^2+10x-1=a^2-(5x-1)^2
=(a-5x-1)(a+5x-1)
98(a-b)^2-18(a+b)^2=2[49(a-b)^2-9(a+b)^2]
=2[(7(a-b))^2-(3(a+b))^2]
=2[(4a-10b)(10a-4b)]
=4[(2a-5b)(5a-2b)]
25x^2-20x+4=(5x-2)^2
by (bi),25(m-1)^2-20m+24=(5(m-1)-2)^2
=(5m-7)^2
(c) part我就懷疑題目出錯/你打錯野,因為我只想到有反證(例如1,2,3,4).所以答唔到....sor