a: \(\frac{\sqrt{x-y}}{\sqrt{x+y}+\sqrt{x-y}}+\frac{x-y}{\sqrt{x^2-y^2}-x+y}\)
\(=\frac{\sqrt{x-y}}{\sqrt{x+y}+\sqrt{x-y}}+\frac{x-y}{\sqrt{\left(x-y\right)}\left(\sqrt{x+y}-\sqrt{x-y}\right)}\)
\(=\frac{\sqrt{x-y}}{\sqrt{x+y}+\sqrt{x-y}}+\frac{\sqrt{x-y}}{\left(\sqrt{x+y}-\sqrt{x-y}\right)}\)
\(=\frac{\sqrt{x-y}\left(\sqrt{x+y}-\sqrt{x-y}+\sqrt{x+y}+\sqrt{x-y}\right)}{\left(\sqrt{x+y}+\sqrt{x-y}\right)\cdot\left(\sqrt{x+y}-\sqrt{x-y}\right)}=\frac{\sqrt{x-y}\cdot2\cdot\sqrt{x+y}}{x+y-x+y}=\frac{2\cdot\sqrt{x^2-y^2}}{2y}=\frac{\sqrt{x^2-y^2}}{y}\)
\(P=\left(\frac{\sqrt{x-y}}{\sqrt{x+y}+\sqrt{x-y}}+\frac{x-y}{\sqrt{x^2-y^2}-x+y}\right)\cdot\frac{x^2y-y^3}{\sqrt{x^2-y^2}}\)
\(=\frac{\sqrt{x^2-y^2}}{y}\cdot\frac{y\left(x^2-y^2\right)}{\sqrt{x^2-y^2}}=x^2-y^2\)
b: P=x^2-y^2
=(x-y)(x+y)
x,y đều là các số nguyên tố lớn hơn 3
=>x và y đều là các số lẻ
=>x=2a+1; y=2b+1
\(x^2-1=\left(2a+1\right)^2-1\)
\(=4a^2+4a+1-1=4a^2+4a=4a\left(a+1\right)\)
Vì a;a+1 là hai số tự nhiên liên tiếp
nên a(a+1)⋮2
=>4a(a+1)⋮4*2
=>4a(a+1)⋮8
=>\(x^2-1\) ⋮8(1)
\(y^2-1=\left(2b+1\right)^2-1\)
\(=4b^2+4b+1-1=4b^2+4b=4b\left(b+1\right)\)
Vì b;b+1 là hai số tự nhiên liên tiếp
nên b(b+1)⋮2
=>4b(b+1)⋮4*2
=>\(y^2-1\) ⋮8(2)
Từ (1),(2) suy ra \(x^2-1-y^2+1\) ⋮8
=>\(x^2-y^2\) ⋮8
=>P⋮8
TH1: x=3a+1; y=3b+1
P=(x-y)(x+y)
=(3a+1-3b-1)(3a+1+3b+1)
=(3a-3b)(3a+3b+2)
=3(a-b)(3a+3b+2)⋮3(3)
TH2: x=3a+2; y=3b+1
P=(x-y)(x+y)
=(3a+2-3b-1)(3a+2+3b+1)
=(3a-3b+1)(3a+3b+3)
=3(a+b+1)(3a-3b+1)⋮3(4)
TH3: x=3a+1; y=3b+2
P=(x-y)(x+y)
=(3a+1-3b-2)(3a+1+3b+2)
=(3a-3b-1)(3a+3b+3)
=3(a+b+1)(3a-3b-1)⋮3(5)
TH4: x=3a+2; y=3b+2
P=(x-y)(x+y)
=(3a+2-3b-2)(3a+2+3b+2)
=(3a-3b)(3a+3b+4)
=3(a-b)(3a+3b+4)⋮3(6)
Từ (3),(4),(5),(6) suy ra P⋮3
mà P⋮8
và ƯCLN(3;8)=1
nên P⋮3*8
=>P⋮24(ĐPCM)
