Question

0,8×*(5y)³+4x³y⁴-xy²(2xy)² при x=-16 y=1/4​

Answer 1

0,8х*125у^3+ 4х^3*у^4 - ху^2*4x^2*у^2= 100ху^3+ 4х^3у^4 - 4х^3у^4 = 100х*у^3= 100*(-16)*(1/4)^3= 100*(-16)*(1/64)= -1600/64= -25