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CRISTINASIMION
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1. [1 supra x+1+ 1 supra 2(x+1)]·x²-1 supra 3
2. Se dă expresia: E(x)= [x supra x²-1 - 1 supra x+1]·(x-1)², x∈R/ [-1, 1]
Vă roog! Dau coroană!!

Răspuns :

Teodoraalbei
1) [1/(x+1)+1/2(x+1)]·(x-1)(x+1) /3
=3 / 2(x+1) ·(x+1)(x-1)=(x-1) / 2

2)E(x)=[x / (x-1)(x+1) -1/(x+1)]·(x-1)²
=1 /(x-1)(x+1)  ·(x-1)²
=(x-1) / (x+1)
MFM
1.[1/(x+1) +1/2(x+1)].(x²-1)/3
aducand la numitorul comun 2(x+1)
[(2+1)/2(x+1)].(x-1)(x+1)/3=
3/2(x+1) ·(x+1)(x-1)/3=
(x-1)/2

2.[x/(x-1)(x+1) -1/(x+1)]·(x-1)²=
numitor comun (x-1)(x+1)
[(x-x+1)/(x-1)(x+1)]·(x-1)²=1/(x-1)(x+1) ·(x-1)²=(x-1)/(x+1)
unde /-fractie

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