B=2π/3-π/4=(8π-3π)/12=5π/12
![\left\{\begin{array}{I}(x+y)^2+3(x+y)+2=0\\ (x-y)^2-5(x-y)+6=0 \end{array}}](https://tex.z-dn.net/?f=+%5Cleft%5C%7B%5Cbegin%7Barray%7D%7BI%7D%28x%2By%29%5E2%2B3%28x%2By%29%2B2%3D0%5C%5C+%28x-y%29%5E2-5%28x-y%29%2B6%3D0+%5Cend%7Barray%7D%7D+)
Решим первое уравнение системы относительно x+y, второе относительно x-y.
1)
![(x+y)^2+3(x+y)+2=0\\ D=9-8=1\\ (x+y)_1=\dfrac{-3+1}{2}=-1\\ (x+y)_2=\dfrac{-3-1}{2}=-2](https://tex.z-dn.net/?f=+%28x%2By%29%5E2%2B3%28x%2By%29%2B2%3D0%5C%5C+D%3D9-8%3D1%5C%5C+%28x%2By%29_1%3D%5Cdfrac%7B-3%2B1%7D%7B2%7D%3D-1%5C%5C++%28x%2By%29_2%3D%5Cdfrac%7B-3-1%7D%7B2%7D%3D-2++)
2)
![(x-y)^2-5(x-y)+6=0\\ D=25-24=1\\ (x-y)_1=\dfrac{5-1}{2}=2\\ (x-y)_2=\dfrac{5+1}{2}=3](https://tex.z-dn.net/?f=+%28x-y%29%5E2-5%28x-y%29%2B6%3D0%5C%5C+D%3D25-24%3D1%5C%5C+%28x-y%29_1%3D%5Cdfrac%7B5-1%7D%7B2%7D%3D2%5C%5C++%28x-y%29_2%3D%5Cdfrac%7B5%2B1%7D%7B2%7D%3D3++)
Получили совокупность четырех систем.
![\left[\begin{array}{I} \left\{\begin{array}{I} x+y=-1 \\ x-y=2 \end{array}} \\ \left\{\begin{array}{I} x+y=-1 \\ x-y=3 \end{array}} \\ \left\{\begin{array}{I} x+y=-2 \\ x-y=2 \end{array}} \\\left\{\begin{array}{I} x+y=-2 \\ x-y=3 \end{array}}\end{array}}](https://tex.z-dn.net/?f=+%5Cleft%5B%5Cbegin%7Barray%7D%7BI%7D+%5Cleft%5C%7B%5Cbegin%7Barray%7D%7BI%7D+x%2By%3D-1++%5C%5C+x-y%3D2+%5Cend%7Barray%7D%7D++%5C%5C+%5Cleft%5C%7B%5Cbegin%7Barray%7D%7BI%7D+x%2By%3D-1++%5C%5C+x-y%3D3+%5Cend%7Barray%7D%7D+%5C%5C+%5Cleft%5C%7B%5Cbegin%7Barray%7D%7BI%7D+x%2By%3D-2++%5C%5C+x-y%3D2+%5Cend%7Barray%7D%7D+%5C%5C%5Cleft%5C%7B%5Cbegin%7Barray%7D%7BI%7D+x%2By%3D-2++%5C%5C+x-y%3D3+%5Cend%7Barray%7D%7D%5Cend%7Barray%7D%7D+++)
Решаем каждую.
1)
![+ \left\{\begin{array}{I} x+y=-1 \\ x-y=2 \end{array}}](https://tex.z-dn.net/?f=+%2B+%5Cleft%5C%7B%5Cbegin%7Barray%7D%7BI%7D+x%2By%3D-1++%5C%5C+x-y%3D2+%5Cend%7Barray%7D%7D+)
![2x=1\\ x=\dfrac{1}{2} \ \Rightarrow \ y= -\dfrac{3}{2}](https://tex.z-dn.net/?f=+2x%3D1%5C%5C+x%3D%5Cdfrac%7B1%7D%7B2%7D+%5C+%5CRightarrow+%5C+y%3D+-%5Cdfrac%7B3%7D%7B2%7D++)
2)
![+ \left\{\begin{array}{I} x+y=-1 \\ x-y=3 \end{array}}](https://tex.z-dn.net/?f=++%2B+%5Cleft%5C%7B%5Cbegin%7Barray%7D%7BI%7D+x%2By%3D-1++%5C%5C+x-y%3D3+%5Cend%7Barray%7D%7D++)
![2x=2\\ x=1 \ \Rightarrow \ y=-2](https://tex.z-dn.net/?f=+2x%3D2%5C%5C+x%3D1+%5C+%5CRightarrow+%5C+y%3D-2+)
3)
![+ \left\{\begin{array}{I} x+y=-2 \\ x-y=2 \end{array}}](https://tex.z-dn.net/?f=+%2B+%5Cleft%5C%7B%5Cbegin%7Barray%7D%7BI%7D+x%2By%3D-2++%5C%5C+x-y%3D2+%5Cend%7Barray%7D%7D+)
![x=0 \ \Rightarrow \ y=-2](https://tex.z-dn.net/?f=+x%3D0+%5C+%5CRightarrow+%5C+y%3D-2+)
4)
![+ \left\{\begin{array}{I} x+y=-2 \\ x-y=3 \end{array}}](https://tex.z-dn.net/?f=+%2B+%5Cleft%5C%7B%5Cbegin%7Barray%7D%7BI%7D+x%2By%3D-2++%5C%5C+x-y%3D3+%5Cend%7Barray%7D%7D+)
![2x=1\\ x=\dfrac{1}{2} \ \Rightarrow \ y=-\dfrac{5}{2}](https://tex.z-dn.net/?f=+2x%3D1%5C%5C+x%3D%5Cdfrac%7B1%7D%7B2%7D+%5C+%5CRightarrow+%5C+y%3D-%5Cdfrac%7B5%7D%7B2%7D+++)
Считаем произведения.
![x_1y_1=\dfrac{1}{2} \cdot (-\dfrac{3}{2})=-\dfrac{3}{4}\\ x_2y_2=1 \cdot (-2)=-2\\ x_3y_3=0 \cdot (-2) =0\\ x_4y_4=\dfrac{1}{2} \cdot (-\dfrac{5}{2})=-\dfrac{5}{4}](https://tex.z-dn.net/?f=+x_1y_1%3D%5Cdfrac%7B1%7D%7B2%7D++%5Ccdot+%28-%5Cdfrac%7B3%7D%7B2%7D%29%3D-%5Cdfrac%7B3%7D%7B4%7D%5C%5C+++x_2y_2%3D1+%5Ccdot+%28-2%29%3D-2%5C%5C+x_3y_3%3D0+%5Ccdot+%28-2%29+%3D0%5C%5C+x_4y_4%3D%5Cdfrac%7B1%7D%7B2%7D+%5Ccdot+%28-%5Cdfrac%7B5%7D%7B2%7D%29%3D-%5Cdfrac%7B5%7D%7B4%7D++++)
Таким образом, наименьшее значение xy=-2
Ответ: -2
Б) (5;-1) подставить место x первое значение в скобках и должно получится второе
<span>Найдите количество целых чисел, принадлежащих множеству значений функции: f(x) =16Log(1/6) (sinx +cosx +3</span>√2) /√2 .
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f(x) =16Log(1/6) (sinx +cosx +3√2) /√2 =16Log(1/6) ( (sinx +cosx)/√2 +3) .
(sinx +cosx) / √2 =(1/√2) *sinx + (1/√2) *cosx) =
cos(π/4) *sinx + sin(π/4) *cosx = sin(π/4+x )
следовательно -1 ≤ (sinx +cosx) /√2 ≤ <span>1 ;
</span>2 ≤ (sinx +cosx) /√2 +3 ≤ 4
т.к. 0 < 1/6 < 1 <span> , то
</span>Log(1/6) 2 ≥ Log(1/6) ( ( sinx +cosx)√2 +3 ) ≥ <span> Log(1/6) </span>4 ;
16*Log(1/6) 2 ≥16* Log(1/6) ( ( sinx +cosx)√2 +3 ) ≥ 16* Log(1/6) 2² ;
32*Log(1/6) 2 ≤ f(x) ≤ 16* Log(1/6) 2 ;
-32*Log(6) 2 ≤ <span>f(x) </span>≤ -16*<span>Log(6) 2 ;
</span><span>-32/(1+Log(2) 3) ≤ f(x) ≤ - 16 / </span><span>(1+Log(2) 3 ) ;
</span> { -12 ; -11; -10 ; -9 ; -8 ; -7 }
ответ : 6 .
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