Question

Verificati daca x=-1 este solutie a inecuatiei:
a)2x+1<-2
b)1+2(x-1)≥4
c)3-x√2>5
d)2x-(3x-1)<0
e)3-2x>1-3(x-1)
f)2≤7+2(x-3)
g)2+x>√3-x-1
h)x+$\frac{2x-1}{5}$≥-1
i)3(2x-3)<2(x-4)-5
Va rooog mult <3

Answer 1

$\displaystyle a).2x+1\ \textless \ -2 \\ 2x\ \textless \ -2-1 \\ 2x\ \textless \ -3 \\ x\ \textless \ - \frac{3}{2} \\ b).1+2(x-1) \geq 4 \\ 1+2x-2 \geq 4 \\ 2x \geq 4-1+2 \\ 2x \geq 5 \\ x \geq \frac{5}{2}$
$\displaystyle c).3-x \sqrt{2} \ \textgreater \ 5 \\ -x \sqrt{2} \ \textgreater \ 5-3 \\ -x \sqrt{2} \ \textgreater \ 2 \\ x\ \textless \ - \frac{2}{ \sqrt{2} } \\ x\ \textless \ -\frac{2 \sqrt{2} }{2} \\ x\ \textless \ - \sqrt{2}$
$d).2x-(3x-1)\ \textless \ 0 \\ 2x-3x+1\ \textless \ 0 \\ -x+1\ \textless \ 0 \\ -x\ \textless \ 0-1 \\ -x\ \textless \ -1 \\ x\ \textgreater \ 1 \\ e).3-2x\ \textgreater \ 1-3(x-1) \\ 3-2x\ \textgreater \ 1-3x+3 \\ -2x+3x\ \textgreater \ 1+3-3 \\ x\ \textgreater \ 1$
$\displaystyle f).2 \leq 7+2(x-3) \\ 2 \leq 7+2x-6 \\ -2x \leq 7-6-2 \\ -2x \leq -1 \\ 2x \geq 1 \\ x \geq \frac{1}{2} \\ g).2+x\ \textgreater \ \sqrt{3} -x-1 \\ x+x\ \textgreater \ \sqrt{3} -1-2 \\ 2x\ \textgreater \ \sqrt{3} -3 \\ x\ \textgreater \ \frac{ \sqrt{3} -3}{2}$
$\displaystyle h).x+ \frac{2x-1}{5} \geq -1 \\ 5x+2x-1 \geq -5 \\ 5x+2x \geq -5+1 \\ 7x \geq -4 \\ x \geq - \frac{4}{7} \\ i).3(2x-3)\ \textless \ 2(x-4)-5 \\ 6x-9\ \textless \ 2x-8-5 \\ 6x-2x\ \textless \ -8-5+9 \\ 4x\ \textless \ -4 \\ x\ \textless \ - \frac{4}{4} \\ x\ \textless \ -1$