1.Care este cel mai mic numar intreg,mai mare decat:a)-61,4; b)12,81. 2.Comparati numerele:a)6,42 si 6,4; b)2supra 3 si 1supra2;c)-5,17si --5,1;d)-41,(71)si -41,71;e) -7,2(1)si -7,(21). 3.Calculati:a)2supra2-5supra6+4supra9-1supra2= b) [0,4-1,2(6)]:[0,2+0,(6)]= c)6supra11[1supra3-3supra10(1supra4-2supra3)]=

1.a)61 b)12,82
2 a) 6,42 > 6,40 b) 2 supra 3 > 1 supra 2(aducem la acelasi numitor amplificand cu 2 si cu 3)
c) 5,17 > 5,10 d) -41,(71) < -41,71 e) -7,2(1) < 7,(21)
3 a) 2/2-5/6+4/9-1/2= amplificam cu 3 prima fractie: 6/6-5/6+4/9-1/2=1/6+4/9-1/2 amplificam primele 2 fractii cu 9 si 6 : 9/54+24/54-1/2=33/54-1/2..amplificam a doua fractie cu 27:33/54-24/54=9/54=prin simplificare cu 9=1/6
restul continua...este usor

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Aivlisask Aivlisask · Answer 2

1.a) 61
b)12,82

2 a) 6,42 > 6,40
b) 2 supra 3 > 1 supra 2(aducem la acelasi numitor amplificand cu 2 si cu 3)
c) 5,17 > 5,10
d) -41,(71) < -41,71
e) -7,2(1) < 7,(21)
3 a) 2/2-5/6+4/9-1/2= (amplificam cu 3 prima fractie) : 6/6-5/6+4/9-1/2=1/6+4/9-1/2 (amplificam primele 2 fractii cu 9 si 6): 9/54+24/54-1/2=33/54-1/2 (amplificam a doua fractie cu 27) :33/54-24/54=9/54=(prin simplificare cu 9) =1/6
b) [tex][ \frac{4}{10} - \frac{126-12}{90} ]:[ \frac{2}{10}+ \frac{6}{9} ]= \frac{36-114}{90} : \frac{18+60}{90} = \frac{-78}{90} * \frac{90}{78} = \frac{-78}{78} =-1[/tex]
c)[tex] \frac{6}{11} [ \frac{1}{3} - \frac{3}{10} ( \frac{1}{4}- \frac{2}{3} )]= \frac{6}{11} [\frac{1}{3} - \frac{3}{10} *( \frac{-5}{12} )]= \frac{6}{11} [ \frac{1}{3}+ \frac{5}{40} ]= \frac{6}{11} [ \frac{40+15}{120} ]= [tex] \frac{6}{11} * \frac{55}{120}= \frac{5}{20} = \frac{1}{4} [/tex]