Question

Решите пожалуйста практическую с развёрнутым ответом

Answer 1

$91^2<91^3\\\\26^4>5^8=25^4\\\\\\2^{3^2}=2^9>2^{2^3}=2^8\\\\\\9^5=3^{10}\ ;\ \ \Big(\dfrac{1}{3}\Big)^{-10}=3^{10}\ \ \to \ \ \ 9^5=\Big(\dfrac{1}{3}\Big)^{-10}\\\\\\(3^{-2})^{-3}=3^6\ ;\ \ \ (3^3)^2=3^6\ \ \to \ \ \ (3^{-2})^{-3}=(3^3)^2\\\\\\27^3=(3^3)^3=3^9>3^6\\\\\\45^2-31^2=(45-31)(45+31)=14\cdot 76>44^2-30^2=(44-30)(44+30)=14\cdot 74$

$\sqrt{\Big(\dfrac{3}{7}\Big)^4}=\Big(\dfrac{3}{7}\Big)^2=\dfrac{9}{49}=\dfrac{36}{196}\ >\ \sqrt{\Big(\dfrac{5}{14}\Big)^4}=\Big(\dfrac{5}{14}\Big)^2=\dfrac{25}{196}\\\\\\\Big(\dfrac{3}{4}\Big)^0=1\ \ ;\ \ \ 1=1\ \ \to \ \ \ \Big(\dfrac{3}{4}\Big)^0=1\\\\\\\Big(\dfrac{3}{4}\Big)^{\pi}<1^{\pi }\ ,\ \ 1^{\pi }=1 \ \ \to \ \ \Big(\dfrac{3}{4}\Big)^{\pi}<1$