Може ли някоя да реши 2 задача

[Гост]

received_626259888485890_edit_1065544675587409.jpg (81.58 KiB) Прегледано 406 пъти

[ammornil]

(a)
[tex]f(x)=3-5x[/tex], [tex]g(x)=\sqrt{5-x}[/tex]

[tex]f(x)-g(x)=3-5x-\sqrt{5-x} \rightarrow DM 5-x\ge0 \Rightarrow x\le5 \Rightarrow x \in (- \infty;5][/tex]

[tex]f(x):g(x)=\frac{3-5x}{\sqrt{5-x}} \rightarrow DM \begin{array}{|l} 5-x\ge0 \\ 5-x\ne0 \end{array} \Leftrightarrow \begin{array}{|l} x\le5 \\ x\ne5 \end{array} \Rightarrow x \in (- \infty;5)[/tex]

Screenshot 2022-01-27 150214.png (38.17 KiB) Прегледано 391 пъти

[ammornil]

(б)
[tex]f(x)=\sqrt[3]{x}[/tex], [tex]g(x)=\sqrt{x}[/tex]

[tex]f(x)-g(x)=\sqrt[3]{x}-\sqrt{x}, DM x\ge0 \Rightarrow x \in [0;+ \infty)[/tex]

[tex]f(x):g(x)=\frac{\sqrt[3]{x}}{\sqrt{x}}, DM \begin{array}{|l} x \ge 0 \\ x \ne 0 \end{array} \Rightarrow x \in (0;+ \infty)[/tex]

Screenshot 2022-01-27 150830.png (28.47 KiB) Прегледано 391 пъти

[ammornil]

(в)
[tex]f(x)=\sqrt{4-x}[/tex], [tex]g(x)=log_{_{5}}{(x^{2}-4)}[/tex]

[tex]f(x)-g(x)=\sqrt{4-x}-log_{_{5}}{(x^{2}-4)} \rightarrow DM \begin{array}{|l} 4-x\ge0 \\x^{2}-4>0 \end{array} \Leftrightarrow \begin{array}{|l} x\le4 \\(x-2)(x+2)>0 \end{array} \Leftrightarrow x \in (- \infty;-2) \cup (2;4][/tex]

[tex]f(x):g(x)=\frac{\sqrt{4-x}}{log_{_{5}}{(x^{2}-4)}}, \rightarrow DM \begin{array}{|l} 4-x\ge0 \\x^{2}-4>0 \\ log_{_{5}}{(x^{2}-4)} \ne 0 \end{array} \Leftrightarrow \begin{array}{|l} x\le4 \\(x-2)(x+2)>0 \\ x^{2}-4 \ne 1 \end{array} \Leftrightarrow[/tex]

[tex]\begin{array}{|l} x\le4 \\(x-2)(x+2)>0 \\x\ne \pm\sqrt{5} \end{array} \Rightarrow x \in (- \infty ;-\sqrt{5}) \cup (-\sqrt{5};-2) \cup (2;\sqrt{5}) \cup (\sqrt{5};4][/tex]

Screenshot 2022-01-27 152312.png (61.9 KiB) Прегледано 391 пъти

[ammornil]

(г)
[tex]f(x)=5^{\frac{x}{x+2}}[/tex], [tex]g(x)=\frac{x-1}{x-2}[/tex]

[tex]f(x)-g(x)=5\Large{^{\frac{x}{x+2}}}\normalsize{-\frac{x-1}{x-2}}, DM \begin{array}{|l} x+2\ne0\\ x-2\ne0 \end{array} \Leftrightarrow \begin{array}{|l} x\ne-2 \\ x\ne2 \end{array} \Rightarrow x \in (- \infty;-2) \cup (-2;2) \cup (2;+ \infty)[/tex]

[tex]f(x):g(x)=\frac{5\Large{^{\frac{x}{x+2}}}}{\normalsize{\frac{x-1}{x-2}}}, DM \begin{array}{|l}x+2\ne0 \\ x-2\ne0 \\ \frac{x-1}{x-2} \ne 0 \end{array} \Leftrightarrow \begin{array}{|l}x\ne-2 \\ x\ne2 \\ x-1 \ne 0 \end{array} \Leftrightarrow \begin{array}{|l}x\ne-2 \\ x\ne2 \\ x \ne 1 \end{array} \Rightarrow x \in (- \infty;-2) \cup (-2;1) \cup (1;2) \cup (2;+ \infty)[/tex]

Screenshot 2022-01-27 153937.png (42.76 KiB) Прегледано 391 пъти

[Гост]

Благодаря много!