Question

Ajutor! Exercițiul 6!​

Answer 1

Răspuns:

x∈(-∞,1)∪[2,+∞)

Explicație pas cu pas:

$\left \{ {{\frac{3-2x}{1-x}}>0 \atop {log_{2}\frac{3-2x}{1-x}}\geq0 }\right.~\left \{ {{\frac{3-2x}{1-x}}>0 } \atop {\frac{3-2x}{1-x}}\geq 1}} \right.~\frac{3-2x}{1-x}}\geq 1,~\frac{3-2x}{1-x}}-1\geq 0,~\frac{3-2x-1+x}{1-x}\geq  0,~\frac{2-x}{1-x}\geq 0,~\\2-x=0,~x=2,~sau \frac{2-x}{1-x}>0,~(2-x)(1-x)>0,$

x∈(-∞,1)∪(2,+∞)∪{2}, deci x∈(-∞,1)∪[2,+∞)