Question

у системі рівнянь {х+у-6ху=-34;ху(х+у)=56 здійснено заміни а=ху,b=x+y.Яку систему рівнянь отримано
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Answer 1

Ответ:

Решаем систему уравнений с помощью замены .

$\left\{\begin{array}{l}\bf x+y-6xy=-34\\\bf xy(x+y)=56\end{array}\right\ \ \ \ \bf a=xy\ ,\ b=x+y\ \ \ \left\{\begin{array}{l}\bf b-6a=-34\\\bf ab=56\end{array}\right\\\\\\\left\{\begin{array}{l}\bf b=6a-34\\\bf a(6a-34)=56\end{array}\right\ \ \left\{\begin{array}{l}\bf b=6a-34\\\bf 6a^2-34a-56=0\end{array}\right\ \ \left\{\begin{array}{l}\bf b=6a-34\\\bf 3a^2-17a-28=0\end{array}\right$  

$\bf 3a^2-17a-28=0\ \ ,\ \ D=b^2-4ac=17^2+4\cdot 3\cdot 28=625\ \ ,\\\\a_1=\dfrac{17-5}{6}=2\ \ ,\ \ \ a_2=\dfrac{17+5}{6}=\dfrac{11}{3}\\\\\\b_1=6\cdot 2-34=-22\ \ \ ,\ \ \ b_2=6\cdot \dfrac{11}{3}-34=22-34=-12$            

Сделаем обратную замену .

$\bf a)\ \left\{\begin{array}{l}\bf xy=2\\\bf x+y=-22\end{array}\right\ \ \left\{\begin{array}{l}\bf x(-x-22)=2\\\bf y=-x-22\end{array}\right\ \ \left\{\begin{array}{l}\bf x^2+22x+2=0\\\bf y=-x-22\end{array}\right\\\\\\x^2+22x+2=0\ \ ,\ \ D/4=11^2-2=119\ \ ,\\\\x_1=-11-\sqrt{119}\ \ ,\ \ x_2=-11+\sqrt{119}\\\\y_1=-33-\sqrt{119}\ \ ,\ \ \ y_2=-33+\sqrt{119}$      

$\bf b)\ \left\{\begin{array}{l}\bf xy=\dfrac{11}{3}\\\bf x+y=-12\end{array}\right\ \ \left\{\begin{array}{l}\bf x(-x-12)=\dfrac{11}{3}\\\bf y=-x-12\end{array}\right\ \ \left\{\begin{array}{l}\bf x^2+12x+\dfrac{11}{3}=0\\\bf y=-x-12\end{array}\right\\\\\\3x^2+36x+11=0\ \ ,\ \ D/4=18^2-3\cdot 11=291\ \ ,\\\\x_1=\dfrac{-18-\sqrt{291}}{3}=-6-\dfrac{\sqrt{291}}{3}\ \ ,\ \ x_2=\dfrac{-18+\sqrt{291}}{3}=-6+\dfrac{\sqrt{291}}{3}\\\\y_1=-18-\dfrac{\sqrt{291}}{3}\ \ ,\ \ \ y_2=-18+\dfrac{\sqrt{291}}{3}$  

$\bf Otvet:\ (-11-\sqrt{119}\ ;\ -33-\sqrt{119}\ )\ ,\ \ (-11+\sqrt{119}\ ;\ -33+\sqrt{119}\ )\ ,\\\\\Big(-6-\dfrac{\sqrt{291}}{3}\ ;\ -18-\dfrac{\sqrt{291}}{3}\ \Big)\ ,\ \Big(-6+\dfrac{\sqrt{291}}{3}\ ;\, -18+\dfrac{\sqrt{291}}{3}\ \Big)\ .$