решите пожалуйста , эти 5 неравенств. Created by meme9. algebra-ru

решите пожалуйста , эти 5 неравенств...

August 2021 1 29 Report

решите пожалуйста , эти 5 неравенств

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YinYang $9*3 \leq \frac{1}{9} * (\frac{1}{27} )^{ \frac{1}{x} } \ \ \ x\neq0!\\
3^{2+1+2} \leq 27^{- \frac{1}{x}} \\
3^5 \leq 3^{- \frac{3}{x} } \\
3 \ \textgreater \ 1 =\ \textgreater \ 5 \leq -\frac{3}{x} \\
 \frac{5x+3}{x} \leq 0 \\
-0.6 \leq x < 0$

$7 \leq ( \frac{1}{49})^{\frac{2}{x}}*(\frac{1}{343})^{\frac{1}{x^2}}\\
7^1 \leq 7^{-\frac{4}{x} -\frac{3}{x^2}} \\
7 \ \textgreater \ 1 =\ \textgreater \ 1 \leq - \frac{4}{x} -\frac{3}{x^2} \\
 \frac{x^2+4x+3}{x^2} \leq 0 \\
-3 \leq x \leq -1
$

$\sqrt{32} *2^{-4x^2} \geq 8^{3x} \\ 2^{2.5-4x^2} \geq 2^{9x} \\ 2 \ \textgreater \ 1 =\ \textgreater \ 2.5-4x^2 \geq 9x \\
8x^2 + 18x - 5 \leq 0 \\
x_{12} = \frac{-9 б \sqrt{81+40}}{8} = \frac{-9б\sqrt{121}}{8} = \frac{-9 б 11}{8} \\
x_1 = -2.5, x_2 = 0.25 \\
-2.5 \leq x \leq 0.25$

$125*(\frac{1}{5})^{x^2} \leq (\frac{1}{25})^{-4x}\\ 5^{3-x^2} \leq 5^{8x} \\ 5 \ \textgreater \ 1 =\ \textgreater \ 3-x^2 \leq 8x \\ x^2+8x-3 \geq 0\\ x_{12} = \frac{-4 б \sqrt{16+3}}{1} = -4 б \sqrt{19} \\ x \leq -4-\sqrt{19}, \ \ x \geq -4+\sqrt{19}$

$log_{\frac{1}{9}}(x+8)+log_{\frac{1}{9}}x \geq -1 ,\ \ \ \ x \ \textgreater \ 0!\\
log_{\frac{1}{9}}(x^2+8x) \geq -1 \\
-0.5log_3(x^2+8x) \geq -1 \\
log_3(x^2+8x) \leq log_39\\
3 \ \textgreater \ 1 =\ \textgreater \ x^2+8x \leq 9 \\
x^2+8x-9 \leq 0$
0 < x ≤ 1

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