1)3^(x+7)-3^(2x)≥0

3^(x+7)≥3^(2x)

x+7≥2x

x≤3.5

Ответ:

1)     [tex]\displaystyle \boldsymbol { D(y)=\{x \in R: x \leq 7\}}[/tex]

2)    [tex]\boldsymbol {D(y)=\{x \in R:x\geq 10\}}[/tex]

Объяснение:

1)

[tex]\displaystyle y=\sqrt{3^{x+7}-9^x} \\\\D(y):\\\\3^{x+7}-9^x\geq 0\\\\3^{x+7}-3^{2x}\geq 0\\\\3^{x+7}\geq 3^{2x}\\\\x+7\geq 2x\\\\x\leq 7[/tex]

[tex]\displaystyle D(y)=\{x \in R: x \leq 7\}[/tex]

2)

[tex]\displaystyle y=\sqrt{\bigg(\frac{1}{125} \bigg)^x-\bigg(\frac{1}{25} \bigg)^{2x-5}}[/tex]

[tex]\displaystyle \bigg(\frac{1}{125} \bigg)^x-\bigg(\frac{1}{25} \bigg)^{2x-5}\geq 0\\\\\\(125)^{-x}-25^{5-2x}\geq 0\\\\\\(5^3)^{-x}-(5^2)^{5-2x}\geq 0\\\\\\5^{-3x}\geq 5^{10-4x}\\\\\\-3x\geq 10-4x\\\\x\geq 10\\\\\\D(y)=\{x \in R:x\geq 10\}[/tex]