Question

Aduceti la forma mai simpla si calculati valoarea expresiei pentru x=1,8 si y=0,6

Answer 1

13a
[x²/(x+y)-x³/(x²+y²+2xy)] :[x/(x+y)-x²/(x²-y²)=
=[x²/(x+y)-x³/(x+y)²] :[x/(x+y)-x²/(x-y)(x+y)=
=[x²(x+y)-x³]/(x+y)² : [x(x-y)-x²]/(x-y)(x+y)=
=x²y/(x+y)² : (-xy)/(x-y)(x+y)=
=x²y/(x+y)² × (x-y)(x+y)/(-xy)=
=x/(x+y) × (x-y)/(-1)=
=x/(x+y) × (y-x)=
=x(y-x)/(x+y)
x=-1,8    y=0,6
x(y-x)/(x+y)=(-1,8)(0,6+1,8)/(-1,8+0,6)=(-4,32)/(-1,2)=3,6

13b
(x+y)/(x+2y) :[x/(x-2y) + y²/(x²-4y²) =
=(x+y)/(x+2y) :[x/(x-2y) + y²/(x-2y)(x+2y) =
=(x+y)/(x+2y) :(x²+2xy+y²)/(x-2y)(x+2y) =
=(x+y)/(x+2y) :(x+y)²/(x-2y)(x+2y) =
=(x+y)/(x+2y) :(x+y)/(x-2y) =
=(x+y)/(x+2y) × (x-2y)/(x+y) =
=(x-2y)/(x+2y)
x=-1,8    y=0,6
(-1,8-2×0,6)/(-1,8+2×0,6)=(-3)/(-0,6)=5


13c
x²/(x²-2y) : [2xy/(x²-4y²) - y/(x+2y)]=
=x²/x(x-2y) : [2xy/(x-2y)(x+2y) - y/(x+2y)]=
=x/(x-2y) : (2xy-xy+2y²)/(x-2y)(x+2y)=
=x/(x-2y) : (xy+2y²)/(x-2y)(x+2y)=
=x/(x-2y) : y(x+2y)/(x-2y)(x+2y)=
=x/(x-2y) : y/(x-2y)=
=x/(x-2y) × (x-2y) /y=
=x/y
x=-1,8    y=0,6
(-1,8)/0,6)=-3

13d
[2xy/(x²-9y²)-y/(x-3y)] :y²/(x²+3xy)=
=[2xy/(x-3y)(x+3y)-y/(x-3y)] :y²/x(x+3y)=
=(2xy-xy-3y²)/(x-3y)(x+3y) × x(x+3y)/y²=
=(xy-3y²)/(x-3y) × x/y²=
=y(x-3y)/(x-3y) × x/y²=
=y/1 × x/y²=
=x/y
x=-1,8    y=0,6
(-1,8)/0,6)=-3