Question

[Nguyễn Anh Thư]

chứng minh rằng

a) \(1000^2+1003^2+1a5^2+1006^2=1002^2+1002^2+1004^2+1007^2\)

b) \(\dfrac{40^2+51^2+91^2}{79^2}=\dfrac{40^4+51^4+91^4}{79^4}\)

HD: a) xét

b) đặt 40=a; 51=b

Answer 1

[The Spryzen123-MC]

Câu a : B-A= 1001² +1002² +1004² +10072 -1000²-1003²-1005²-1006²

=(1001²-1000²)+(1002²-1003²) (1004²-1005²)+(1007²-1006²)

HĐT số 3= (1001-1000)*(1001+1000)+(1002-1003)*(1002+1003)+(1004-1005)*(1004+1005)+(1007-1006)*(1007+1006)

=2001 -2005-2009+2013=0

vậy A=B