Question

Exercițiul 1 va rog ofer coroana​

Answer 1

$\displaystyle a)~\left|\begin{array}{ccc}x-4&1-3x\\5&-3\end{array}\right|-\frac{2}{3} \left|\begin{array}{ccc}8&-2\\2&7\end{array}\right|=\frac{3}{2} \left|\begin{array}{ccc}5&1,(6)\\3&-1\end{array}\right|\\\\(x-4) \cdot (-3)-(1-3x) \cdot 5-\frac{2}{3} \cdot (8 \cdot 7-(-2)\cdot 2)=\frac{3}{2} \cdot \Bigg(5 \cdot (-1)-\frac{15}{9} \cdot 3\Bigg)\\\\-3x+12-5+15x-\frac{2}{3} \cdot (56+4)=\frac{3}{2} \cdot \Bigg(-5-\frac{45}{9}\Bigg)$

$\displaystyle 12x+7-\frac{2}{3} \cdot 60=\frac{3}{2} \cdot (-5-5)\\\\ 12x+7-\frac{120}{3} =\frac{3}{2} \cdot (-10)\\\\12x+7-40=-\frac{30}{2} \\\\ 12x-33=-15 \\\\ 12x=-15+33\\\\12x=18\\\\ x=\frac{18}{12}\Rightarrow x=\frac{3}{2} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~S_1=\Bigg\{\frac{3}{2}\Bigg\}$

$b)~\left|\begin{array}{ccc}y+4&-y-5&y+1\\1&y-1&3\\y+2&-1&y\end{array}\right|=\left|\begin{array}{ccc}2y+1&-1\\1&2y\end{array}\right|\\\\\\ (y+4) \cdot (y-1) \cdot y+(y+1)\cdot 1\cdot(-1)+(-y-5)\cdot 3 \cdot (y+2)-\\\\-(y+1)\cdot (y-1) \cdot (y+2)-(-y-5)\cdot 1 \cdot y-(y+4)\cdot 3\cdot (-1)=\\\\=(2y+1) \cdot 2y-(-1) \cdot 1\\\\\Big(y^2-y+4y-4\Big) \cdot y+(y+1) \cdot (-1)+(-3y-15) \cdot (y+2)-\\\\-\Big(y^2-y+y-1\Big) \cdot (y+2)-(-y-5)\cdot y-(3y+12) \cdot (-1)=\\\\=4y^2+2y+1$

$\displaystyle \Big(y^2+3y-4\Big) \cdot y-y-1-3y^2-6y-15y-30-\Big(y^2-1\Big)\cdot (y+2)+\\\\+y^2+5y+3y+12=4y^2+2y+1\\\\ y^3+3y^2-4y-y-1-3y^2-6y-15y-30-y^3-2y^2+y+2+y^2+5y+\\\\+3y+12=4y^2+2y+1\\\\ -y^2-17y-17=4y^2+2y+1\\\\ -y^2-4y^2-17y-2y-17-1=0\\\\ -5y^2-19y-18=0 \Big|\cdot (-1)\\\\ 5y^2+19y+18=0$

$\displaystyle \Delta=19^2 -4 \cdot 5 \cdot 18=361-360=1 > 0\\\\ y_1=\frac{-19-\sqrt{1} }{2\cdot 5} =\frac{-19-1}{10}=-\frac{20}{10} =-2 \\\\ y_2=\frac{-19+\sqrt{1} }{2\cdot 5} =\frac{-19+1}{10}=-\frac{18}{10} =-\frac{9}{5} \\\\ S_2=\Bigg\{-2;-\frac{9}{5} \Bigg\}$