![2^{2x}-3\cdot2^x+2 \leq 0
\\\
(2^{x})^2-3\cdot2^x+2 \leq 0
\\\
(2^x-2)(2^x-1) \leq 0
\\\
2^x-2=0 \Rightarrow 2^x=2 \Rightarrow x=1
\\\
2^x-1=0 \Rightarrow 2^x=1 \Rightarrow x=0
\\\
x\in[0;1]
\\\
x_{max}=1\in Z](https://tex.z-dn.net/?f=2%5E%7B2x%7D-3%5Ccdot2%5Ex%2B2+%5Cleq+0%0A%5C%5C%5C%0A%282%5E%7Bx%7D%29%5E2-3%5Ccdot2%5Ex%2B2+%5Cleq+0%0A%5C%5C%5C%0A%282%5Ex-2%29%282%5Ex-1%29+%5Cleq+0%0A%5C%5C%5C%0A2%5Ex-2%3D0+%5CRightarrow+2%5Ex%3D2+%5CRightarrow+x%3D1%0A%5C%5C%5C%0A2%5Ex-1%3D0+%5CRightarrow+2%5Ex%3D1+%5CRightarrow+x%3D0%0A%5C%5C%5C%0Ax%5Cin%5B0%3B1%5D%0A%5C%5C%5C%0Ax_%7Bmax%7D%3D1%5Cin+Z "2^{2x}-3\cdot2^x+2 \leq 0
\\\
(2^{x})^2-3\cdot2^x+2 \leq 0
\\\
(2^x-2)(2^x-1) \leq 0
\\\
2^x-2=0 \Rightarrow 2^x=2 \Rightarrow x=1
\\\
2^x-1=0 \Rightarrow 2^x=1 \Rightarrow x=0
\\\
x\in[0;1]
\\\
x_{max}=1\in Z")
Ответ: 1
0
\\\
125\cdot5^{3-x}-5^{x-3}>0
\\\
\frac{125}{5^{x-3}} -5^{x-3}>0
\\\
125-5^{2x-6}>0
\\\
5^{2x-6}<125
\\\
5^{2x-6}<5^3
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2x-6<3
\\\
2x<9
\\\
x<4,5
\\\
x_{max}=4\in Z" alt="250\cdot5^{3-x}-2\cdot5^{x-3}>0
\\\
125\cdot5^{3-x}-5^{x-3}>0
\\\
\frac{125}{5^{x-3}} -5^{x-3}>0
\\\
125-5^{2x-6}>0
\\\
5^{2x-6}<125
\\\
5^{2x-6}<5^3
\\\
2x-6<3
\\\
2x<9
\\\
x<4,5
\\\
x_{max}=4\in Z" align="absmiddle" class="latex-formula">
Ответ: 4