(\frac{1}{25})^{16-x}\\ 5^{-x^2-2x} > 5^{-32+2x}\\ -x^2-2x>2x-32\\ -x^2-4x+32>0\\
(-8;4)" alt="1) 3^{-4}=\frac{1}{81}\\
(\frac{4}{7})^{-1}=\frac{7}{4}\\
27^{\frac{1}{3}}+49^{\frac{1}{2}}=3+7=10\\
(1+5^{\frac{2}{3}})(1-5^{\frac{2}{3}}+5^{\frac{4}{3}})=25+1=26\\
\\
3)\\
2^{0.5+3x}=2^{-1}\\
0.5+3x=-1\\
3x=-1.5\\
x=-0.5\\
\\
2^{2x}+4*2^{x}-12=0\\
2^x=a\\
a^2+4a-12=0\\
D=8^2\\
a=2\\
2^x=2\\
x=1
\\[tex](\frac{1}{5})^{x^2+2x}>(\frac{1}{25})^{16-x}\\ 5^{-x^2-2x} > 5^{-32+2x}\\ -x^2-2x>2x-32\\ -x^2-4x+32>0\\
(-8;4)" align="absmiddle" class="latex-formula">
5)
на концах отрезков
![f(x)=\frac{3x^{\frac{2}{3}}}{2}-\frac{x^3}{3}\\
f(0)=0\\
f(8)=-\frac{494}{3}\\
\\
f'(x)=(\frac{3x^{\frac{2}{3}}}{2}-\frac{x^3}{3})'= \frac{1}{\sqrt[3]{x}}-x^2\\
f'(x)=0\\
x=1\\](https://tex.z-dn.net/?f=+f%28x%29%3D%5Cfrac%7B3x%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%7D%7B2%7D-%5Cfrac%7Bx%5E3%7D%7B3%7D%5C%5C%0A+f%280%29%3D0%5C%5C%0A+f%288%29%3D-%5Cfrac%7B494%7D%7B3%7D%5C%5C%0A%5C%5C%0Af%27%28x%29%3D%28%5Cfrac%7B3x%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%7D%7B2%7D-%5Cfrac%7Bx%5E3%7D%7B3%7D%29%27%3D+%5Cfrac%7B1%7D%7B%5Csqrt%5B3%5D%7Bx%7D%7D-x%5E2%5C%5C%0Af%27%28x%29%3D0%5C%5C%0A+x%3D1%5C%5C "f(x)=\frac{3x^{\frac{2}{3}}}{2}-\frac{x^3}{3}\\
f(0)=0\\
f(8)=-\frac{494}{3}\\
\\
f'(x)=(\frac{3x^{\frac{2}{3}}}{2}-\frac{x^3}{3})'= \frac{1}{\sqrt[3]{x}}-x^2\\
f'(x)=0\\
x=1\\")
поподает поставим будет 7/6
значит наибольшее 7/6 наименьшее -494/3
6)
![f(-2)=3^{-2}-2=\frac{1}{9}-2= -\frac{17}{9}\\
f(7)=3^{7}-2=2185\\
\\
\\
f(-2)=\sqrt[3]{-2+1}=-1\\
f(7)=\sqrt[3]{7+1}=2\\](https://tex.z-dn.net/?f=f%28-2%29%3D3%5E%7B-2%7D-2%3D%5Cfrac%7B1%7D%7B9%7D-2%3D+-%5Cfrac%7B17%7D%7B9%7D%5C%5C%0Af%287%29%3D3%5E%7B7%7D-2%3D2185%5C%5C%0A%5C%5C%0A%5C%5C%0Af%28-2%29%3D%5Csqrt%5B3%5D%7B-2%2B1%7D%3D-1%5C%5C%0Af%287%29%3D%5Csqrt%5B3%5D%7B7%2B1%7D%3D2%5C%5C%0A "f(-2)=3^{-2}-2=\frac{1}{9}-2= -\frac{17}{9}\\
f(7)=3^{7}-2=2185\\
\\
\\
f(-2)=\sqrt[3]{-2+1}=-1\\
f(7)=\sqrt[3]{7+1}=2\\")
Рисунки внизу второй задачи соотвественно