Odpowiedź:
16
[tex]5\cdot 2^{\sqrt{x} }-3\cdot2^{\sqrt{x} -1}\geq 56\qquad D= < 0,+\infty)\\\displaystyle 5\cdot 2^{\sqrt{x} }-\frac{3}{2} \cdot2^{\sqrt{x} }\geq 56\\\frac{7}{2} \cdot2^{\sqrt{x} }\geq 56/\cdot \frac{2}{7} \\2^{\sqrt{x} }\geq 16\\2^{\sqrt{x} }\geq2^4\quad \Rightarrow \quad \sqrt{x} \geq 4/^2\\x\geq 16\\\underline{16}[/tex]
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2025 KUDO.TIPS - All rights reserved.
Odpowiedź:
16
[tex]5\cdot 2^{\sqrt{x} }-3\cdot2^{\sqrt{x} -1}\geq 56\qquad D= < 0,+\infty)\\\displaystyle 5\cdot 2^{\sqrt{x} }-\frac{3}{2} \cdot2^{\sqrt{x} }\geq 56\\\frac{7}{2} \cdot2^{\sqrt{x} }\geq 56/\cdot \frac{2}{7} \\2^{\sqrt{x} }\geq 16\\2^{\sqrt{x} }\geq2^4\quad \Rightarrow \quad \sqrt{x} \geq 4/^2\\x\geq 16\\\underline{16}[/tex]