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Решение задания приложено
B3 2(x+3)½=-x
(x+3)½=-x/2 возводим обе части в квадрат
x+3=x²/4 / *4
x²-4x-12=0
x₁=6
x₂=-2
ОТВЕТ: 6
Обозначим x=количество литров 2-го раствора,тогда имеем :
0,2·5+0,6x=0.4(5+10+x);
1+0.6x=2+4+0.4x;
0.2x=5;
x=5/0.2=25литров;
Решение
<span><span>По условию </span>x1 /x2= 2</span>
Сделаем уравнение
приведенным
x^2+(3k-1)/ (k^2-5k+3) x+2/(k^2-5k+3) =0
по теореме Виета
p = (3k-1)/ (k^2-5k+3)
<span>x1 + x2 = - p = - (3k-1)/ (k^2-5k+3)</span>
2*x2 +x2 = - (3k-1)/ (k^2-5k+3)
<span>3*<span>x2 = - (3k-1)/ (k^2-5k+3)
</span></span><span>X2 = - (3k-1)/ 3(k^2-5k+3) (1)</span>
<span>q = 2/(k^2-5k+3) </span>
<span>x1*x2 = q = 2/(k^2-5k+3) ;
</span><span>2*x2 *x2 = 2/(k^2-5k+3) ;</span>
<span>X2^2 = 1/(k^2-5k<span>+3) (2)</span></span>
Подставляем (1) в (2)
<span> <span>( - (3k-1)/ 3)^2 = (k^2-5k+3)
</span></span><span>(1-3k)^2 /9 = (k^2-5k+3)
</span><span>(1-3k)^2 = 9k^2 -45k +27</span>
<span>1 -6k +9k^2 = 9k^2 -45k
+27</span>
<span>45k – 6k = 27 -1</span>
<span>39k = 26</span>
<span>K = 26/39 = 2/3</span>
Проверка
<span><span>Подставим </span><span>k= 2/3 </span><span>в исходное уравнение
</span></span>
((2/3)^2-5*(2/3)+3)x^2+(3*(2/3)-1)x+2=0
Преобразуем
<span>X^2 +9x +18 = 0</span>
<span>D = 9^2 -4*1*18 = 9 ; √D = +/- 3</span>
<span>X = 1/2<span><span>* ( - 9 +/- 3)
</span></span></span><span>X1 = - 6
</span><span>X<span>2 = </span>-3</span>
ПРОВЕРКА
<span>X1 / X2 = <span> - 6 / - 3 = 2</span></span>
<span><span>ОТВЕТ </span>k = 2<span>/3</span></span>
