Question

решите...
1) (xy)² при x = -6 и y = -1,5; x = 0 и y = -23.
2) (x+y)⁴ при x = 0,7 и y = 0,3; x = -11 и y = -6.
3)(y-x)³ при x = -14 и y = -10; x =0,9 и y = 1,1

Answer 1

Здравствуй!

#1

$boxed{;bf (frac{x}{y} )^{2} esli x=-6, y=-1.5 ili x=0, y=-23 ;}\\left[begin{array}{ccc}(frac{-6}{-1.5})^2=4^2=16end{array}right] \\left[begin{array}{ccc}(frac{0}{-23})^2=0^2=0end{array}right]$

#2

$boxed{;bf (x+y )^{4} esli x=0.7, y=0.3 ili x=-11, y=-6 ;}\\left[begin{array}{ccc}(0.7+0.3)^4=1^4=1end{array}right] \\left[begin{array}{ccc}(-11-6)^4=-17^4=83521end{array}right]$

#3

$boxed{;bf (y-x)^{3} esli x=-14, y=-10 ili x=0.9, y=1.1 ;}\\left[begin{array}{ccc}(-10-(-14))^4=(-10+14)^4=4^4=256end{array}right] \\left[begin{array}{ccc}(1,1-0,9)^4=0,2^4=0,0016end{array}right]$

Удачи в учёбе!