Question

Решите пожалуйста Алгебра

Answer 1

$5 ^{x-1} \leq 25\\\\5 ^{x-1} \leq 5 ^{2}\\\\x-1 \leq 2\\\\x \leq 3$
x∈ (- ∞ : 3]


$( \frac{1}{2} ) ^{ x^{2} - x-4} \geq \frac{1}{2}\\\\ x^{2} -3x-4 \leq 1\\\\ x^{2} -3x-5 \leq 0\\\\ x^{2} -3x-5=0\\\\D=(-3) ^{2} -4*1*(-5)=9+20=29\\\\ x_{1} = \frac{3+ \sqrt{29} }{2}\\\\ x_{2}= \frac{3- \sqrt{29} }{2} \\\\(x- \frac{3+ \sqrt{29} }{2})(x- \frac{3- \sqrt{29} }{2}) \leq 0$
x ∈ $[ \frac{3- \sqrt{29} }{2} ; \frac{3+ \sqrt{29} }{2} ]$