Question

((343x^3+125y^3)/(7x-5y))*((49x^2-70xy+25y^2)/(49x^2-35xy+25y^2)) при x=√31 y=√47
Помогите пожалуйста.

Answer 1

$frac{343x^3+125y^3}{7x-5y}* frac{49x^2-70xy+25y^2}{49x^2-35xy+25y^2}= frac{(7x+5y)(49x^2-35xy+25y^2) }{7x-5y}* frac{(7x-5y)^2}{49x^2-35xy+25y^2}$=$(7x+5y)(7x-5y)$=$49x^2-25y^2$