Question

срочно!! даю 88 баллов

Answer 1

Ответ:

1)     $\displaystyle \boldsymbol { D(y)=\{x \in R: x \leq 7\}}$

2)    $\boldsymbol {D(y)=\{x \in R:x\geq 10\}}$

Объяснение:

1)

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

$\displaystyle D(y)=\{x \in R: x \leq 7\}$

2)

$\displaystyle y=\sqrt{\bigg(\frac{1}{125} \bigg)^x-\bigg(\frac{1}{25} \bigg)^{2x-5}}$

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