Partea întreaga a lui x ∈ R :
Partea facționala a lui x ∈ R :
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[tex]\bf a)\; \dfrac{3}{5} \cdot 4 =\dfrac{3 \cdot 4}{5} = \dfrac{12}{5} = 2,4[/tex]
[tex]\bf 2 < 2,4 < 3 \implies \Bigg[ \dfrac{12}{5} \Bigg] = \boxed{\bf2}[/tex]
[tex]\bf \Bigg\{\dfrac{12}{5} \Bigg\} = \dfrac{12}{5} - \Bigg[ \dfrac{12}{5} \Bigg] = \dfrac{12}{5} - 2 = \boxed{\bf\dfrac{2}{5 } }[/tex]
[tex]\bf \;[/tex]
[tex]\bf b)\; 8 \cdot \dfrac{2}{3} = \dfrac{2\cdot 8}{3} = \dfrac{16}{3} =5,(3)[/tex]
[tex]\bf 5 < 5,(3) , 6 \implies \Bigg[\dfrac{16}{3} \Bigg] = \boxed{\bf 5 }[/tex]
[tex]\bf \Bigg\{ \dfrac{16}{3} \Bigg\} = \dfrac{16}{3} - \Bigg[\dfrac{16}{3}\Bigg] = \dfrac{16}{3} - 5 = \boxed{\bf \dfrac{1}{3} }[/tex]
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[tex]\bf c)\; \dfrac{4}{7} \cdot 3 = \dfrac{4 \cdot 3}{7} =\dfrac{12}{7} =1,(714285)[/tex]
[tex]\bf 1 < 1,(714285) < 2 \implies \Bigg[ \dfrac{12}{7} \Bigg] = \boxed{\bf 1 }[/tex]
[tex]\bf \Bigg\{\dfrac{12}{7} \Bigg\} = \dfrac{12}{7} - \Bigg[\dfrac{12}{7} \Bigg] = \dfrac{12}{7} -1= \boxed{\bf \dfrac{5}{7} }[/tex]
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[tex]\bf d)\; \dfrac{12}{25} \cdot 5 = \dfrac{12 \cdot 5 }{25} ^{(5} = \dfrac{12}{5} =2,4[/tex]
[tex]\bf 2<2,4<3 \implies \Bigg[\dfrac{12}{5} \Bigg] = \boxed{\bf 2 }[/tex]
[tex]\bf \Bigg\{\dfrac{12}{5} \Bigg\} = \dfrac{12}{5} - \Bigg[\dfrac{12}{5} \Bigg] = \dfrac{12}{5} -2 = \boxed{\bf \dfrac{2}{5} }[/tex]
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[tex]\bf e)\; 8 \cdot \dfrac{7}{16} = \dfrac{8 \cdot 7}{16} ^{(8} = \dfrac{7}{2} = 3,5[/tex]
[tex]\bf 3 < 3,5 < 4 \implies \Bigg[ \dfrac{7}{2} \Bigg] = \boxed{\bf 3 }[/tex]
[tex]\bf \Bigg\{\dfrac{7}{2} \Bigg\} = \dfrac{7}{2} - \Bigg[ \dfrac{7}{2} \Bigg] =\dfrac{7}{2} - 3 = \boxed{\bf \dfrac{1}{2}}[/tex]
[tex]\bf \;[/tex]
[tex]\bf f)\; \dfrac{11}{30} \cdot 12 = \dfrac{11\cdot 12}{30} ^{(6} = \dfrac{22}{5} = 4,4[/tex]
[tex]\bf 4<4,4<5 \implies \Bigg[\dfrac{22}{5} \Bigg] = \boxed{\bf 4 }[/tex]
[tex]\bf \Bigg\{ \dfrac{22}{5} \Bigg\} = \dfrac{22}{5} - \Bigg[ \dfrac{22}{5} \Bigg] = \dfrac{22}{5} - 4 = \boxed{\bf \dfrac{2}{5} }[/tex]
[tex]\bf \;[/tex]
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