Question

11 класс, задача повышенной сложности. Вообще-то ничего сложного, но ответ у меня не совпал.

Answer 1

$\log_{x-3}(x^2-4x)^2\leq4$

ограничения: $\displaystyle\left\{{{x^2-4x\neq0}\atop{\left\{{{x-3\ \textgreater \ 0}\atop{x-3\neq1}}\right}}\right\to\left\{{{\left\{{{x\neq0}\atop{x\neq4}}\right}\atop{\left\{{{x\ \textgreater \ 3}\atop{x\neq4}}\right}}\right$, следовательно, $x\in(3;4)(4;+\infty)$

решение: 

$\log_{x-3}(x^2-4x)^2\leq4;~\log_{x-3}(x^2-4x)^2-4\leq0;~\log_{x-3}(x^2-4x)^2-\\-\log_{x-3}(x-3)^4\leq0;~(x-3-1)[(x^2-4x)^2-(x-3)^4]\leq0;~\\(x-4)[(x^2-4x)^2-(x^2-6x+9)^2]\leq0;~\\(x-4)(x^2-4x-x^2+6x-9)(x^2-4x+x^2-6x+9)\leq0;~\\(x-4)(2x-9)(2x^2-10x+9)\leq0$

до конца раскладываем на множители: 

$(x-4)(x-\frac{9}{2})(x-\frac{5+\sqrt{7}}{2})(x-\frac{5-\sqrt{7}}{2})\leq0$

теперь метод интервалов, который дарит нам ответ на это неравенство: $x\in(3;\frac{5+\sqrt{7}}{2})(4;\frac{9}{2}]$

Answer 2

ОДЗ
{x-3>0⇒x>3
{x-3≠1⇒x≠4
{x²-4x≠0⇒x(x-4)≠0⇒x≠0 U x≠4
x∈(3;4) U (4;∈(3;4)
(x²-4x)²≥(x-3)^4
(x²-4x)²-(x-3)^4≥0
(x²-4x)²-(x²-6x+9)²≥0
(x²-4x-x²+6x-9)(x²-4x+x²-6x+9)≥0
(2x-9)(2x²-10x+9)≥0
2x-9=0⇒x=4,5
2x²-10x+9=0
D=100-72=28
x1=(10-2√7)/4=2,5-0,5√7 U x2=2,5+0,5√7
             _                          +                                 _                    +
                               ///////////////////////////////                          ////////////////////
----------------[2,5+0,5√7]-----(3)-------[2,5+0,5√7]------(4)--------[4,5]--------------
                                            \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
x∈(3;2,5+0,5√7]
2)x∈(4;∞)
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\                                //////////////////////////////////////
          _                                  +                                  _                    +          
----------------[2,5+0,5√7]-----(3)-------[2,5+0,5√7]------(4)--------[4,5]-----------
                                                                                  \\\\\\\\\\\\\\\\\\\\\\\\\\\\\
x∈(4;4,5]
Ответ x∈(3;2,5+0,5√7] U (4;4,5]

Answer 3

Task Explanation

The task involves a C++ code snippet and requires understanding the behavior of the program. Let's analyze the code and determine the output.

Code Analysis

The given C++ code snippet involves a base class `Parent` and a derived class `Derived`. The `Derived` class overrides the `print` method from the `Parent` class. There are two instances of the `main` function, each creating an object of the `Derived` class and calling the `print` method.

class Parent {
public:
    virtual void print() {
        std::cout << " " << std::endl;
    }
};

class Derived : public Parent {
public:
    virtual void print(int x) {
        std::cout << " " << std::endl;
    }
};

int main() {
    Derived *derived = new Derived;
    derived->print();
    return 0;
}

int main() {
    Parent *derived = new Derived;
    derived->print();
    return 0;
}

Output Analysis

The output of the program depends on the concept of function overloading and virtual functions in C++. The first `main` function creates an object of type `Derived` and calls the `print` method. The second `main` function creates an object of type `Parent` (pointing to a `Derived` object) and calls the `print` method.

The output will be determined by the behavior of virtual functions and function overloading in C++.

Conclusion

The output of the given C++ code will be based on the behavior of virtual functions and function overloading in C++. The specific output will depend on the rules governing these concepts in C++.

Note: The provided code contains two `main` functions, which is not valid in C++. It seems to be a typographical error. If the second `main` function is intended to be a separate program, it should be in a different file.

Let's discuss the output based on the analysis of the code.