URGENT!!!! EXERCIȚIUL 6

Răspuns:

Explicație pas cu pas:

[tex]a = \frac{4}{3}-\frac{4}{3} *[\frac{4}{3} -(0,(3))^2]:4\frac{8}{9} =[/tex]

[tex]= \frac{4}{3}-\frac{4}{3} *[\frac{4}{3} -(\frac{3}{9} )^2]:\frac{4*9 + 8}{9} =[/tex]

[tex]= \frac{4}{3}-\frac{4}{3} *[\frac{4}{3} -(\frac{1}{3} )^2]:\frac{36 + 8}{9} =[/tex]

[tex]= \frac{4}{3}-\frac{4}{3} *(\frac{4}{3} -\frac{1}{3^2}):\frac{44}{9} =[/tex]

[tex]= \frac{4}{3}-\frac{4}{3} *(\frac{4*3}{3*3} -\frac{1}{9})*\frac{9}{44} =[/tex]

[tex]= \frac{4}{3}-\frac{4}{3} *(\frac{12}{9} -\frac{1}{9})*\frac{9}{44} =[/tex]

[tex]= \frac{4}{3}-\frac{4}{3} *\frac{12-1}{9}*\frac{9}{44} =[/tex]

[tex]= \frac{4}{3}-\frac{4}{3} *\frac{11}{9}*\frac{9}{44} =[/tex]

[tex]= \frac{4}{3}-\frac{4}{3} *\frac{11}{9}*\frac{9}{4*11} =[/tex]

[tex]= \frac{4}{3}-\frac{1}{3} =[/tex]

[tex]= \frac{4-1}{3} =[/tex]

[tex]= \frac{3}{3} =[/tex]

[tex]= 1[/tex] ∈ N

[tex]\it(0,(3))^2=\Big(\dfrac{3}{9}\Big)^2=\dfrac{\ 9^{(9}}{81}=\dfrac{1}{9}[/tex]

[tex]\it n=\dfrac{4}{3}-\dfrac{4}{3}\cdot\Big(\dfrac{^{3)}4}{\ 3}-\dfrac{1}{9}\Big):\dfrac{44}{9}=\dfrac{4}{3}-\dfrac{4}{3}\cdot\dfrac{11}{9}\cdot\dfrac{9}{44}=\dfrac{4}{3}-\dfrac{1}{3}=\dfrac{3}{3}=1\in\mathbb{N}[/tex]