Question

Potęgi 1 liceum zadanie w załączniku

Answer 1

(49/4)^1/2*4*1/14+12-8 = 7/2*2/7+4 = 5

redukuje się potęga 2 z 1/2 z zewnętrznych nawisów i zostaje
7+24^1/2-7-24^1/2 = 0

^ oznacza potęgę

Answer 2

1.
(49/4)^(1/2)(125/1000)^(-2/3) 14^(-1)+72^(1/2)2^(1/2)-
(128/100)^3:(64/100)^3=
=(7/2)(5/10)^(-2)(1/14)+(72 razy2)^(1/2)-[(128/100)razy(100/64)]^3=
=(7/2)(100/25)(1/14)+144^(1/2)-(128/64)^3=
=1+12-2^3=13-8=5

gdzie np.(49/4)^(1/2) oznacza 49/4 do potegi 1/2

2.
[(7+24^(1/2)-(7-24^(1/2)]^2=[7+24^(1/2)]-2[(7+24^(1/2)]^(1/2)[7-24^(1/2)]^(1/2)+[7-24^(1/2)]=14-2{[7+24^(1/2)][7-24^(1/2)]}^(1/2)=
14-2[49-24]^(1/2)=14-2(25)^(1/2)=14-2 razy5=14-10=4

Answer 3

W zał