Question

$\sqrt{3x} \: + 2 \leqslant 2x + \sqrt{3}$
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dau coroană ✨✨✨​

Answer 1

$\it \sqrt3x+2\leq2x+\sqrt3 \Rightarrow 2-\sqrt3\leq2x-\sqrt3x \Rightarrow 2-\sqrt3\leq x(2-\sqrt3)|_{:(2-\sqrt3)}\\ \\ \Rightarrow 1\leq x \Rightarrow x\geq1 \Rightarrow x\in[1,\ \ \infty)$

Answer 2

Răspuns:

√3x-2x≤√3-2 ;   x(√3-2)≤√3-2 | *(-1)   ;   x(2-√3)≥2-√3 ;    x≥1 ;   x∈[1;∞)

Explicație pas cu pas: