Question

Rezolvati ecuatiile:
a) 1\4{1\4[1\4(1\4 ori x +1) +4] +1} +4 =5
b) 12 intregi si 1\3 : {[ 2 intregi si 3\4: (1 intreg si 2\3+ 1 intreg si 7\8 ori x)] 8\11+1 inreg si 2\3}=5
c) 0,(1x)+0,(2x)+...+0, (9x) = x
Rezolvari complete!!!

Answer 1

[tex] \frac{1}{4}*\{ \frac{1}{4}*[ \frac{1}{4}*( \frac{x}{4}+1)+4]+1\}+4=5 \\ \\ \frac{1}{4}*\{\frac{1}{4}*[ \frac{1}{4}*( \frac{x}{4}+1)+4]+1\}=1 \\ \\ \frac{1}{4}*[ \frac{1}{4}*( \frac{x}{4}+1)+4]+1=1: \frac{1}{4} \\ \\ \frac{1}{4}*[ \frac{1}{4}*( \frac{x}{4}+1)+4]+1=4 \\ \\ \frac{1}{4}*[\frac{1}{4}*( \frac{x}{4}+1)+4]=3 \\ \\ \frac{1}{4}*( \frac{x}{4}+1)+4=3: \frac{1}{4} \\ \\ \frac{1}{4}*( \frac{x}{4}+1)+4=12 \\ \\ \frac{1}{4}*( \frac{x}{4}+1)=8 \\ \\ \frac{x}{4}+1=8: \frac{1}{4} \\ \frac{x}{4}+1=32 \\[/tex]
[tex]\frac{x}{4}=31 \\ \\ \boxed{x=124}[/tex]



[tex]0,(1x)+0,(2x)+.......+0,(9x)=x \\ \\ \frac{1x}{99}+ \frac{2x}{99}+........+ \frac{9x}{99}= \frac{99x}{99} \\ \\ 10+x+20+x+.....+90+x=99x \\ \\ 9x+450=99x \\ \\ 90x=450 \\ \\ \boxed{x=5}[/tex]