Question

a)[1,1(6)+0,(36)+1,25]÷367 supra 396
b)[0,1(36)+0,(432)+0,2×0,(135):0,11]÷221 supra 407

Answer 1

A) 1,1(6)=(116-11)/90=105/90=7/6;
0,(36)=36/99=4/11;
1,25=125/100=5/4.
Inlocuim [(7/6)+(4/11)+(5/4)]:(367/396)= (367/132):(367/396)=3;
(In paranteza dreapta numitorul comun este 132)
B) 0,1(36)=(136-1)/990=3/22;
0,(432)=432/999=16/37;
0,2=2/10=1/5;
0,(135)=135/999=5/37;
0,11=11/100.
Inlocuim [(3/22)+(16/37)+(1/5)*(5/37):(11/100)]:(221/407)= (663/814)*(407/221)=3/2;
(Numitorul la paranteza dreapta este 814, dupa ce efectuam inmultirea si impartirea).

Answer 2

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