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641.v

TEKST ZADATKA

Uprosti sledeći izraz:

(y+12y−2+62y2−2−y+32y+2)⋅4y2−43\left( \frac{y+1}{2y-2} + \frac{6}{2y^2-2} - \frac{y+3}{2y+2} \right) \cdot \frac{4y^2-4}{3}
REŠENJE ZADATKA

Prvo, određujemo uslove definisanosti izraza. Imenioci ne smeju biti jednaki nuli:

2y−2≠0  ⟹  y≠12y+2≠0  ⟹  y≠−12y2−2≠0  ⟹  y≠±12y-2 \neq 0 \implies y \neq 1 \\ 2y+2 \neq 0 \implies y \neq -1 \\ 2y^2-2 \neq 0 \implies y \neq \pm 1

Faktorišemo imenioce u zagradi kako bismo našli najmanji zajednički sadržalac:

2y−2=2(y−1)2y2−2=2(y2−1)=2(y−1)(y+1)2y+2=2(y+1)2y-2 = 2(y-1) \\ 2y^2-2 = 2(y^2-1) = 2(y-1)(y+1) \\ 2y+2 = 2(y+1)

Zapisujemo izraz sa faktorisanim imeniocima:

(y+12(y−1)+62(y−1)(y+1)−y+32(y+1))⋅4y2−43\left( \frac{y+1}{2(y-1)} + \frac{6}{2(y-1)(y+1)} - \frac{y+3}{2(y+1)} \right) \cdot \frac{4y^2-4}{3}

Proširujemo razlomke u zagradi do najmanjeg zajedničkog imenioca, koji je 2(y−1)(y+1): 2(y-1)(y+1) :

(y+1)(y+1)+6−(y+3)(y−1)2(y−1)(y+1)⋅4y2−43\frac{(y+1)(y+1) + 6 - (y+3)(y-1)}{2(y-1)(y+1)} \cdot \frac{4y^2-4}{3}

Množimo polinome u brojiocu:

(y2+2y+1)+6−(y2+2y−3)2(y−1)(y+1)⋅4y2−43\frac{(y^2+2y+1) + 6 - (y^2+2y-3)}{2(y-1)(y+1)} \cdot \frac{4y^2-4}{3}

Sređujemo brojilac oduzimanjem i sabiranjem sličnih monoma:

y2+2y+1+6−y2−2y+32(y−1)(y+1)⋅4y2−43=102(y−1)(y+1)⋅4y2−43\frac{y^2 + 2y + 1 + 6 - y^2 - 2y + 3}{2(y-1)(y+1)} \cdot \frac{4y^2-4}{3} = \frac{10}{2(y-1)(y+1)} \cdot \frac{4y^2-4}{3}

Skraćujemo razlomak u zagradi sa 2 2 i faktorišemo brojilac drugog razlomka:

5(y−1)(y+1)⋅4(y2−1)3=5(y−1)(y+1)⋅4(y−1)(y+1)3\frac{5}{(y-1)(y+1)} \cdot \frac{4(y^2-1)}{3} = \frac{5}{(y-1)(y+1)} \cdot \frac{4(y-1)(y+1)}{3}

Množimo razlomke i skraćujemo zajedničke faktore (y−1)(y+1) (y-1)(y+1) (uz uslov y≠±1 y \neq \pm 1 ):

5⋅4(y−1)(y+1)3(y−1)(y+1)=203\frac{5 \cdot 4(y-1)(y+1)}{3(y-1)(y+1)} = \frac{20}{3}