Question

Ajutati-ma va rog cu problema 10 !!!!

Answer 1

$a) \frac{x}{12} = \frac{\frac{1}{3}+\frac{1}{4}}{7}  \\\frac{x}{12}= \frac{\frac{4}{12}+\frac{3}{12}}{7}\\\frac{x}{12}= \frac{\frac{7}{12}}{7}\\ \frac{x}{12}= \frac{7}{12}:{7}\\\frac{x}{12}= \frac{7}{12}*\frac{1}{7}\\\frac{x}{12}= \frac{1}{12}\\ Produsul\ miezilor\ este\ egal\ cu\ produsul\ extremilor\\12*x=12\ ->x=\frac{12}{12}=1$

$c) \frac{3}{8} = \frac{\frac{75}{100}+\frac{27}{4}}{x+1}\\    \frac{3}{8} = \frac{\frac{3}{4}+\frac{27}{4}}{x+1}\\    \frac{3}{8} = \frac{\frac{30}{4}}{x+1}\\    \frac{3}{8} = \frac{15}{2}:(x+1)\\    \frac{3}{8} = \frac{15}{2}*\frac{1}{x+1}\\    \frac{3}{8} = \frac{15}{2*(x+1)}\\    2*3*(x+1)=8*15\\    6*(x+1)=120\\    x+1=120:6=20\\    x=20-1=19$

$e) \frac{2^{51}-2^{50}-2^{49}}{8^{16}} =\frac{31x}{992} \\   \frac{2^{49}*(2^{2}-2-1)}{2^{16*3}} =\frac{x}{32} \\    \frac{2^{49}*(4-3)}{2^{48}} =\frac{x}{32}\\    \frac{2^{49}}{2^{48}} =\frac{x}{32}\\    2=\frac{x}{32}\\    64=x$

$d) \frac{\frac{2}{10}+\frac{2}{9}}{2x+1} = \frac{1}{45}\\   \frac{\frac{18}{90}+\frac{20}{90}}{2x+1} = \frac{1}{45}\\    \frac{\frac{38}{90}}{2x+1} = \frac{1}{45}\\    \frac{38}{90}:(2x+1) = \frac{1}{45}\\     \frac{38}{90}*\frac{1}{2x+1} = \frac{1}{45}\\      \frac{38}{90*(2x+1)} = \frac{1}{45}\\    38*45=90*(2x+1)\\     \frac{38*45}{90} =2x+1\\ 19=2x+1\\2x=19-1=18\\x=9$

$b) \frac{\frac{1}{5}-\frac{1}{5}:5}{4} = \frac{x}{75}\\     \frac{\frac{1}{5}-\frac{1}{5}*\frac{1}{5}}{4} = \frac{x}{75}\\     \frac{\frac{5}{25}-\frac{1}{25}}{4} = \frac{x}{75}\\     \frac{\frac{4}{25}}{4} = \frac{x}{75}\\     \frac{4}{25}*\frac{1}{4} = \frac{x}{75}\\     \frac{1}{25}=\frac{x}{75}\\     25x=75\\     x=3$

$\frac{x}{14+4}=\frac{8}{x}\\ \frac{x}{18}=\frac{8}{x}\\x^{2}=18*8=144\\x=12$