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Proszę o pomoc w rozwiązaniu zadań. W pierwszym wyszło mi x1=1 i x2= \(-\frac{13}{44}\) a powinno wyjść x1=1 i x2 =0 . Pozostałych nie potrafię zrobić



a)\(5^{2x+1} +3 * 10^{x} -2 ^{2x+1} \le 0\)
b)\(\frac{ 15^{x} }{ 14^{x}- 15^{x} } \le 14\)
c)\(( \frac{13}{31} )^{13x ^{2}-31x } \ge ( \frac{31}{13}) ^{31x ^{2}-13 }\)

A w drugim i trzecim co powinno wyjść...

mmmaaamm pisze: c)\(( \frac{13}{31} )^{13x ^{2}-31x } \ge ( \frac{31}{13}) ^{31x ^{2}-13 }\)

\(( \frac{13}{31} )^{13x ^{2}-31x } \ge ( \frac{31}{13}) ^{31x ^{2}-13 }\\
( \frac{13}{31} )^{13x ^{2}-31x } \ge ( \frac{13}{31}) ^{-31x ^{2}+13 }\\
13x^2-31x\leq -31x^2+13\\
44x^2-31x-13\leq 0\\
\sqrt{\Delta}=57\\
x_1=-\frac{13}{44}\\
x_2=1\\
x\in \left\langle -\frac{13}{44},1\right\rangle\)

a)
\(5^{2x+1} +3 * 10^{x} -2 ^{2x+1} \le 0/:10^x
5( \frac{5}{2})^x+3-2 \cdot ( \frac{2}{5} )^x \le 0
( \frac{5}{2})^x=t>0
5t+3-2 \frac{1}{t} \le 0
(5t^2+3t-2)t \le 0
t=0
\Delta =49
t_1=\frac{2}{5}
t_2=-1
t \in (- \infty ,-1> \cup<0,\frac{2}{5}>
t \in (- \infty ,-1> \cup<0,\frac{2}{5}> \wedge t>0 \Rightarrow t \in (0,\frac{2}{5}>
( \frac{5}{2})^x=\frac{2}{5}=( \frac{5}{2})^{-1}
x=-1
x \le -1\)

b)\(\frac{15^{x}}{14^{x}-15^{x}} \le 14 \Leftrightarrow \frac{1}{( \frac{14}{15})^{x}-1 } \le 14\)
\(( \frac{14}{15})^{x}=t \wedge t>0 \wedge t \neq 1\)
\(\frac{1}{t-1} \le 14 \to \frac{-14t+15}{t-1} \le 0 \to (-14t+15)(t-1) \le 0 \wedge t>0 \wedge t \neq 1\)

\(t \in (0;1) \cup [ \frac{15}{14};+ \infty )\)
\(\begin{cases} ( \frac{14}{15})^{x}<1 \\( \frac{14}{15})^{x}>0 \end{cases} \vee ( \frac{14}{15})^{x} \ge \frac{15}{14}\)

\(x>0 \vee x \le -1\)

\(odp.x \in (- \infty ;-1] \cup (0;+ \infty )\)