ta có x³+y³+z³=(x+y+z)(x²+y²+z²-xy-yz-xz)+3xyz(hằng đẳng thức)

Theo đề bài ->x+y+z=0

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\) =>  (yz + xz + xy) / xyz = 0  => yz + zx + xy = 0 

Ta có : x² + 2yz = x² + yz + yz = x² + yz - zx - xy = x.(x - z) - y.(x - z) = (x - y).(x - z)

Tương tự, y² + 2xz = y² + xz + xz = y² + xz - xy - yz = y(y - x) + z(x - y) = (x - y)(z - y)

; z² + 2xy = (x - z).(y - z)

Vậy \(A=\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(x-y\right)\left(z-y\right)}+\frac{xy}{\left(x-z\right)\left(y-z\right)}\)

\(A=\frac{yz\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}-\frac{xz\left(x-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}+\frac{xy\left(x-y\right)}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}\)

\(A=\frac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{yz\left(y-z\right)-xz\left(x-y+y-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)

\(A=\frac{\left(yz-xz\right)\left(y-z\right)+\left(xy-xz\right)\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)

cảm ơn

thấy 43 là số lẻ => 2 số a và b phải có 1 số là số chẵn nguyên tố 
=> số chẵn nguyên tố đó chỉ có thể là 2 
=> a = 2 , b= 41