15 задание, срочно нужно помогите пожалуйста

$7^x-7 \neq 0 \\ 7^x \neq 7 \\ x \neq 1$

$7^x-4 \neq 0 \\ 7^x \neq 4 \\ x \neq log_{7}4$

$y=7^x$

$\frac{2}{y-7} - \frac{5}{y-4} \geq 0 \\ y \neq 7 \\ y \neq 4$

$\frac{2(y-4)-5(y-7)}{(y-7)(y-4)} \geq 0 \\ \\ \frac{2y-8-5y+35}{(y-7)(y-4)} \geq 0 \\ \\ = \frac{-3y+27}{(y-7)(y-4) } \geq 0$
$\frac{-3(y-9)}{(y-7)(y-4)} \geq 0 \\ \\ \frac{y-9}{(y-7)(y-4)} \leq 0 \\ \\ (y-9)(y-7)(y-4) \leq 0$
y=9 y=7 y=4
- + - +
-------- 4 ----------- 7 ------------ 9 -------------
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y∈(-∞; 4)U(7; 9]

1) $7^x\ \textless \ 4 \\ \\ x\ \textless \ log_{7}4$
2) $7\ \textless \ 7^x \leq 9 \\ 1\ \textless \ x\ \textless \ log_{7}9$

x∈(-∞; log₇4)U(1; log₇9]