Question

Алгебра
номер 7.77 под номером 1 и 3​

Answer 1

$(-\infty; 0);$

$(-2; +\infty);$

Объяснение:

$1) \quad \bigg (\dfrac{2}{3} \bigg )^{x}+\bigg (\dfrac{2}{3} \bigg )^{x-1}2,5;$

$\bigg (\dfrac{2}{3} \bigg )^{x} \cdot \bigg (1+\bigg (\dfrac{2}{3} \bigg )^{-1} \bigg )2,5;$

$\bigg (\dfrac{2}{3} \bigg )^{x} \cdot \bigg (1+\dfrac{3}{2} \bigg )2,5;$

$\bigg (\dfrac{2}{3} \bigg )^{x} \cdot 2,52,5;$

$\bigg (\dfrac{2}{3} \bigg )^{x}1;$

$\bigg (\dfrac{2}{3} \bigg )^{x}\bigg (\dfrac{2}{3} \bigg )^{0};$

$\bigg (\dfrac{3}{2} \bigg )^{-x}\bigg (\dfrac{3}{2} \bigg )^{0};$

$-x0;$

$x$

$x \in (-\infty; 0);$

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$3) \quad \bigg (\dfrac{4}{3} \bigg )^{x+1}-\bigg (\dfrac{4}{3} \bigg )^{x}\dfrac{3}{16};$

$\bigg (\dfrac{4}{3} \bigg )^{x} \cdot \bigg (\dfrac{4}{3}-1 \bigg )\dfrac{3}{16};$

$\bigg (\dfrac{4}{3} \bigg )^{x} \cdot \dfrac{1}{3}\dfrac{3}{16} \quad \bigg | \quad \cdot 3$

$\bigg (\dfrac{4}{3} \bigg )^{x}\dfrac{9}{16};$

$\bigg (\dfrac{4}{3} \bigg )^{x}\bigg (\dfrac{3}{4} \bigg )^{2};$

$\bigg (\dfrac{4}{3} \bigg )^{x}\bigg (\dfrac{4}{3} \bigg )^{-2};$

$x-2;$

$x \in (-2; +\infty);$