Kopertina e librit Matematika 10 - 11: Pjesa I

Zgjidhja e ushtrimit 15

Zgjidhja e ushtrimit 15 të mësimit Vlerësim 5 në librin Matematika 10 - 11: Pjesa I nga shtëpia botuese Botime Pegi me autorë Steve Fearnley, June Haighton, Steve Lomax, Peter Mullarkey, James Nicholson dhe Matt Nixon.

Pyetja

Ktheni në thyesa numrat dhjetorë periodikë të mëposhtëm. Paraqitni secilën thyesë në trajtën e saj më të thjeshtë.

  1. 0.18‾0.\overline{18}
  2. 0.26‾0.2\overline6
  3. 0.22‾0.\overline{22}
  4. 0.465‾0.4\overline{65}
  5. 0.227‾0.2\overline{27}
  6. 0.3514‾0.35\overline{14}

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Zgjidhja

  1. x=0.18‾⇒100x=18.18‾⇒100x−x=18.18‾−0.18‾⇒99x=18⇒x=1899=211x = 0.\overline{18} \rArr100x = 18.\overline{18} \rArr 100x-x=18.\overline{18} - 0.\overline{18} \rArr 99x=18 \rArr x=\dfrac{\cancel{18}}{\cancel{99}}=\boxed{\dfrac{2}{11}}
  2. x=0.26‾⇒100x=26.6‾⇒100x−10x=26.6‾−2.6‾⇒90x=24⇒x=2490=415x=0.2\overline6 \rArr 100x=26.\overline6 \rArr 100x-10x=26.\overline6-2.\overline6 \rArr90x=24 \rArr x=\dfrac{24}{90}=\boxed{\dfrac{4}{15}}
  3. x=0.22‾⇒x=0.2‾⇒10x=2.2‾⇒10x−x=2.2‾−0.2‾⇒9x=2⇒x=29x=0.\overline{22} \rArr x=0.\overline2 \rArr 10x=2.\overline2 \rArr 10x-x=2.\overline2-0.\overline2 \rArr 9x=2 \rArr x=\boxed{\dfrac{2}{9}}
  4. x=0.465‾⇒1000x=465.65‾⇒1000x−10x=465.65‾−4.65‾⇒990x=461⇒x=461990x=0.4\overline{65} \rArr1000x=465.\overline{65} \rArr 1000x-10x=465.\overline{65}-4.\overline{65} \rArr990x=461 \rArr x=\boxed{\dfrac{461}{990}}
  5. x=0.227‾⇒1000x=227.27‾⇒1000x−10x=227.27‾−2.27‾⇒990x=225⇒x=225990=522x=0.2\overline{27} \rArr 1000x=227.\overline{27} \rArr 1000x-10x=227.\overline{27}-2.\overline{27} \rArr 990x=225 \rArr x=\dfrac{225}{990}=\boxed{\dfrac{5}{22}}
  6. x=0.3514‾⇒10000x=3514.14‾⇒10000x−100x=3514.14‾−35.14‾⇒9900x=3479⇒x=34799900x=0.35\overline{14} \rArr 10000x=3514.\overline{14} \rArr 10000x-100x=3514.\overline{14}-35.\overline{14} \rArr 9900x=3479 \rArr x=\boxed{\dfrac{3479}{9900}}