Question

1.Verificati daca numarul rational 1 supra 3 este solutie pentru ecuatia:
a.x:16 supra 15=5 supra 8
b.3x(x+1 supra 5 )=2

2.Rezolvati ecuatia [1,(3)x supra 2 - 2 intregi si 1 pe 4]:0,6=x supra 9

Answer 1

1)   
$a) \\ x: \frac{16}{15} = \frac{5}{8} \\ \\ x= \frac{5}{8}~\cdot~\frac{16}{15}=\frac{5~\cdot~16}{8~\cdot~15}=\frac{1~\cdot~2}{1~\cdot~3}=\boxed{\frac{2}{3}} \neq \boxed{\frac{1}{3}} \\ \\ b) \\ 3x( \frac{x+1}{5})=2 \\ \\ \frac{3x(x+1)}{5}=2 \\ 3x(x+1)=2~\cdot~5 \\ 3x^2+3x=10 \\ 3x^2+3x-10=0 \\ \text{Verificam daca }\; \frac{1}{3} \;\; \text{este solutie a ecuatiei: }$

$\\ \\ 3( \frac{1}{3} )^2 + 3~\cdot~ \frac{1}{3}-10 = \\ \\ =3~\cdot~ \frac{1}{9} + 3~\cdot~ \frac{1}{3}-10 = \\ \\ = \frac{3}{9} + \frac{3}{3}-10= \\ \\ = \frac{1}{3} +1-10= \frac{1}{3} -9=\frac{1}{3} - \frac{27}{3} = - \frac{26}{3} \neq 0 \\ \\ =\ \textgreater \ ~~\frac{1}{3} ~\text{nu este solutie nici pentru ecuatia a) nici pentru ecuatia b).}$


2)
$[ \frac{1,(3)x}{2} - 2 \frac{1}{4}]:0,6= \frac{x}{9} \\ \\ ([\frac{ \frac{4}{3} x}{2} - \frac{9}{4}]: \frac{3}{5} = \frac{x}{9} \\ \\ \frac{ 2x}{3} - \frac{9}{4}= \frac{x}{9}~\cdot~\frac{3}{5} \\ \\ \frac{ 2}{3}x - \frac{9}{4}= \frac{x}{15} \\ \\ \frac{ 2}{3}x -\frac{1}{15}x =\frac{9}{4} \\ \\ (\frac{10}{15} -\frac{1}{15})x=\frac{9}{4} \\ \\ \frac{3}{5}x=\frac{9}{4} ~~~ =>~~x = \frac{9}{4}~\cdot~\frac{5}{3}=\boxed{\frac{15}{4}}$