Прошу решите срочно нужно

1) [latex]log_ \frac{1}{3} (x-1) \geq -2 \\ \left \{ {{x-1 \leq 9} \atop {x-1\ \textgreater \ 0}} \right. \\ \left \{ {{x\leq10} \atop {x\ \textgreater \ 1}} \right. [/latex] x ∈ (1; 10] 2) [latex]13^{2x+1} -13^x-12=0 \\ 13^x=t \\ 13t^2-t-12=0 \\ D=1+13*48=625 \\ x_1=- \frac{12}{13} \\ x_2=1 \\ \\ 13^x=- \frac{12}{13} \\ 13^x=1[/latex] x ∈ ∅ x = 0 x = 0 3) cos α = 0,8;  0 < α < π/2 [latex] \left \{ {{|sin \alpha|= \sqrt{1-0,8^2} } \atop {0\ \textless \ \alpha \ \textless \ \frac{ \pi }{2} }} \right. [/latex] sin α = 0,6 tg α = sin α / cos α = 0,6/0,8 = 3/4 ctg α = 1 / tg α = 4/3 4) [latex] \sqrt{x-1}=x-3 \\ \left \{ {{x-1=x^2-6x+9} \atop {x-3 \geq 0}} \right.\\ \left \{ {{x^2-7x+10=0} \atop {x \geq 3}} \right.\\ \left \{ {{x_1=2;x_2=5} \atop {x \geq 3}} \right. \\ x=5 [/latex] 5) [latex] \sqrt{2+x-x^2} \ \textgreater \ -1[/latex] 2 + x - x² ≥ 0 x² - x - 2 ≤ 0 x² - x - 2 = 0 По теореме Виета: x₁ = -1, x₂ = 2 x ∈ [-1; 2]