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a fost răspuns

2+x√3>5
2-3(x-1)<-1
4-x<2(x-1)
x-4>2(x-3)-1
1-x>x-3+√2
Va rogggg

Răspuns :

Deni00
1) [tex]2+x\sqrt{3}\ \textgreater \ 5|-2\\ x\sqrt{3}\ \textgreater \ 5-2\\ x\sqrt{3}\ \textgreater \ 3 \\ x\ \textgreater \ \frac{3}{\sqrt{3}}\\ x\ \textgreater \ \frac{3\sqrt{3}}{3}}\\ x\ \textgreater \ \sqrt{3}\\ x \in (\sqrt{3},+\infty) [/tex]
2) 4-x<2(x-1)
4-x<2x-2
4+2<2x+x
3x>6
x>2
x∈(2,+∞)
3) x-4>2(x-3)-1
x-4>2x-6-1
x-4>2x-7
x-2x>-7+4
-x>-3|*(-1)
x<3
x∈(-∞,3)
4)
[tex]1-x\ \textgreater \ x-3+\sqrt{2}\\ -x-x\ \textgreater \ -3+\sqrt{2}-1\\ -2x\ \textgreater \ -4+\sqrt{2}|*(-1) \\ 2x\ \textless \ 4-\sqrt{2}|:2\\ x\ \textless \ \frac{4-\sqrt{2}}{2}\\ x \in(-\infty, \frac{4-\sqrt{2}}{2})[/tex]

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