Помогите пожалуйста.математика 10 класс,помогите хоть с одним

[latex]\left\{\begin{array}{c}x-4\neq0,\\x^2-8x+16\geq0,\\x-1\neq0,\\x^2-2x+1\geq0,\\5x-6-x^2\ \textgreater \ 0,\\45x-9x^2-54\geq0;\end{array}\right. \ \left\{\begin{array}{c}x-4\neq0,\\(x-4)^2\geq0,\\x-1\neq0,\\(x-1)^2\geq0,\\5x-6-x^2\ \textgreater \ 0,\\5x-x^2-6\geq0;\end{array}\right. \ \left\{\begin{array}{c}(x-4)^2\neq0,\\(x-1)^2\neq0,\\5x-x^2-6\ \textgreater \ 0;\end{array}\right.[/latex] [latex]\left\{\begin{array}{c}x-4\neq0,\\x-1\neq0,\\x^2-5x+6\ \textless \ 0;\end{array}\right. \ \left\{\begin{array}{c}x\neq4,\\x\neq1,\\(x-2)(x-3)\ \textless \ 0;\end{array}\right. \ \left\{\begin{array}{c}x\neq4,\\x\neq1,\\2\ \textless \ x\ \textless \ 3;\end{array}\right. \\ 2\ \textless \ x\ \textless \ 3;[/latex] [latex]\frac{5\sqrt{x^2-8x+16}}{x-4}-\frac{11\sqrt{x^2-2x+1}}{x-1}+\frac{\sqrt{45x-9x^2-54}}{\sqrt{5x-6-x^2}}=\\=\frac{5\sqrt{(x-4)^2}}{x-4}-\frac{11\sqrt{(x-1)^2}}{x-1}+\frac{3\sqrt{5x-6-x^2}}{\sqrt{5x-6-x^2}}=\\=\frac{5|x-4|}{x-4}-\frac{11|x-1|}{x-1}+\frac{3\sqrt{5x-6-x^2}}{\sqrt{5x-6-x^2}}=\frac{-5(x-4)}{x-4}-\frac{11(x-1)}{x-1}+\frac{3\sqrt{5x-6-x^2}}{\sqrt{5x-6-x^2}}=\\=-5-11+3=-13.[/latex] [latex]\sqrt[4]{x^2-2|x|+1}=\sqrt[4]{(|x|-1)^2}=\sqrt{||x|-1|}, \\ 1) \ y=\sqrt{x}, \\ 2) \ y=-\sqrt{x}, \\ 3) \ y=-\sqrt{|x|}, \\ 4) \ y=-\sqrt{|x|-1}, \\ 5) \ y=-\sqrt{||x|-1|}.[/latex]