Question

sa se rezolve exercitiile

Answer 1

$8) \quad L = \lim_{n \to 2} \dfrac{x-2}{x^4-16} \overset{ \boxed{\frac{0}{0}} }= \lim_{n \to 2} \dfrac{(x-2)^'}{(x^4-16)^'} = \lim_{n \to 2} \dfrac{1}{4x^3} = \\ \dfrac{1}{4\cdot2^3} }=\\ \\ = \dfrac{1}{4\cdot8} = \dfrac{1}{32} \Rightarrow \boxed{ f) corect }$


$9)\quad log__{\big{x-1}}} 1+ \sqrt{x-1} + \sqrt{2-x} \geq0 \\ \\ \boxed{ Conditii de existenta: } \\ \\ \boxed{1} \quad x-1 \ \textgreater \ 0 \Rightarrow x\ \textgreater \ 1 \Rightarrow x \in (1, +\infty) \\ \\ \boxed{2} \quad x-1 \neq1 \Rightarrow x \neq 2 \\ \\ \boxed{3} \quad 1+ \sqrt{x-1}+ \sqrt{2-x} \ \textgreater \ 0 \Rightarrow \\ \Rightarrow \sqrt{x-1}+ \sqrt{2-x} \ \textgreater \ -1 \\ \sqrt{x-1}\geq0 \\ \sqrt{2-x} \geq 0 \\ \\ \Rightarrow \sqrt{x-1}+ \sqrt{2-x} \ \textgreater \ -1,\quad \forall x\in \mathbb_{R}$

$\boxed{4} \quad x-1 \geq 0 \Rightarrow x \geq 1 \\ \\ \boxed{5} \quad 2-x \geq 0 \Rightarrow -x \geq -2 \Rightarrow x \geq 2$

$Din \boxed{1} \cap \boxed{2} \cap \boxed{3} \cap \boxed{4} \cap \boxed{5} \Rightarrow x\in (1,2) = D$


$log__{\big{x-1}}} 1+ \sqrt{x-1} + \sqrt{2-x} \geq0 \Rightarrow \\ \\ \Rightarrow log__{\big{x-1}}} 1+ \sqrt{x-1} + \sqrt{2-x} \geq log__{\big{x-1}}} 1 \\ \\ Din D \Rightarrow 1\ \textless \ x\ \textless \ 2 \Big|-1 \Rightarrow 0\ \textless \ x-1\ \textless \ 1 \quad( baza subunitara) \\ \Rightarrow functia logaritmica este strict descrescatoare \\ \\ \Rightarrow 1+ \sqrt{x-1} + \sqrt{2-x} \leq 1 \Rightarrow \sqrt{x-1} + \sqrt{2-x} \leq 1-1 \Rightarrow \\ \Rightarrow \sqrt{x-1} + \sqrt{2-x} \leq 0 \\ \\ Dar, stim ca \sqrt{x-1} + \sqrt{2-x} \geq0$

$\Rightarrow \sqrt{x-1} + \sqrt{2-x} = 0 \\ \\ \Rightarrow \sqrt{x-1}=0 si \sqrt{2-x} = 0 \Rightarrow x=1 si x=2 \\ \\ \Rightarrow x \in \big\{1, 2\big\} \\ \\ Din \big\{1, 2\big\} \cap D \Leftrightarrow \big\{1, 2\big\} \cap \big(1,2\big) \Rightarrow \boxed{S = \O} \\ \\ Se pare ca niciuna dintre variante nu este corecta!$