←Vrati se na zadatke1. ZadatakRiješi jednadžbu (1+12006):20072005+1x−(2−0.2⋅103):43=1−1:0.2−0.050.8−0.65.\left(1+\frac1{2006}\right):\frac{2007}{2005}+\frac1x-\left(2-0.2\cdot\frac{10}3\right):\frac43=1-1:\frac{0.2-0.05}{0.8-0.65}.(1+20061):20052007+x1−(2−0.2⋅310):34=1−1:0.8−0.650.2−0.05.Prikaži rješenjeRješenjeSređivanjem dane jednadžbe dobivamo 20072006⋅20052007+1x−(2−23)⋅34=1−1:0.150.15\frac{2007}{2006}\cdot\frac{2005}{2007}+\frac1x-\left(2-\frac23\right)\cdot\frac34=1-1:\frac{0.15}{0.15}20062007⋅20072005+x1−(2−32)⋅43=1−1:0.150.15, odnosno 20052006+1x−43⋅34=1−1:1\frac{2005}{2006}+\frac1x-\frac43\cdot\frac34=1-1:120062005+x1−34⋅43=1−1:1, odakle je 20052006+1x−1=1−1\frac{2005}{2006}+\frac1x-1=1-120062005+x1−1=1−1. Sada je 1x=1−20052006=12006\frac1x=1-\frac{2005}{2006}=\frac1{2006}x1=1−20062005=20061. Konačno, slijedi da je rješenje dane jednadžbe x=2006x=2006x=2006.Pomakni formulu lijevo ili desno.
