Страница 69 номер 260, ГДЗ по алгебре за 9 класс к учебнику Мерзляка
Найдите нули функции: 1) \(f(x) = 0{,}2x + 3\); 2) \(g(x) = 35 - 2x - x^2\); 3) \(\varphi(x) = \sqrt{x + 3}\); 4) \(h(x) = \frac{x^2 - x - 6}{x + 3}\); 5) \(f(x) = x^3 - 4x\); 6) \(f(x) = x^2 + 1\).
1) \(f(x) = 0{,}2x + 3\)
\(f(x) = 0 \Rightarrow 0{,}2x + 3 = 0 \Rightarrow 0{,}2x = - 3 \Rightarrow x = - 15\)
2) \(g(x) = 35 - 2x - x^{2}\)
\(g(x) = 0 \Rightarrow - x^{2} - 2x + 35 = 0 \Rightarrow\)
\(\Rightarrow x^{2} + 2x - 35 = 0 \Rightarrow (x - 5)(x + 7) = 0 \Rightarrow x_{1} = 5;\ x_{2} = - 7\)
3) \(\varphi(x) = \sqrt{x + 3}\)
\(\sqrt{x + 3} = 0 \Rightarrow x + 3 = 0 \Rightarrow x = - 3\)
4) \(h(x) = \frac{x^{2} - x - 6}{x + 3}\)
\(\begin{cases} x^{2} - x - 6 = 0 \\ x + 3 \neq 0 \end{cases}\ \Rightarrow \begin{cases} (x + 2)(x - 3) = 0 \\ x \neq - 3 \end{cases}\ \Rightarrow\)
\(\Rightarrow \begin{cases} x = - 2;\ x = 3 \\ x \neq - 3 \end{cases}\ \Rightarrow x_{1} = - 2;\ x_{2} = 3\)
5) \(f(x) = x^{3} - 4x\)
\(x^{3} - 4x = 0 \Rightarrow x\left( x^{2} - 4 \right) = 0 \Rightarrow\)
\(\Rightarrow x(x - 2)(x + 2) = 0 \Rightarrow x_{1} = - 2;\ x_{2} = 0;\ x_{3} = 2\)
6) \(f(x) = x^{2} + 1\)
\(x^{2} + 1 > 0\) при \(x \in ( - \infty; + \infty)\), т.е. функция не имеет нулей.
Ответ: 1) \(x = - 15\); 2) \(x_{1} = 5;\ x_{2} = - 7\); 3) \(x = - 3\); 4) \(x_{1} = - 2;\ x_{2} = 3\); 5) \(x_{1} = - 2;\ x_{2} = 0;\ x_{3} = 2\); 6) функция не имеет нулей.
