Question

Rezolvați subiectul 1 și subiectul 2​

Answer 1

$\displaystyle SUBIECTUL~I\\\\ 1.~~\Bigg(2-\Bigg(\frac{1}{3} \Bigg)^2\Bigg):\frac{17}{9} =\Bigg(2-\frac{1^2}{3^2} \Bigg):\frac{17}{9} =\Bigg(2-\frac{1}{9} \Bigg):\frac{17}{9} =\frac{18-1}{9} :\frac{17}{9} =\\\\\\=\frac{17}{9} \cdot \frac{9}{17} =1$

$\displaystyle 2.~~f:\mathbb{R}\rightarrow \mathbb{R},~f(x)=4x+a,~a\in\mathbb{R}\\\\ f(2)-f(-2)=4 \cdot 2+a-(4 \cdot (-2)+a)=8+a-(-8+a)=\\\\=8+a+8-a=16$

$\displaystyle 3.~~3^{x^2+3}=3^{4x}\Rightarrow x^2+3=4x \Rightarrow x^2-4x+3=0\\\\ \Delta=(-4)^2-4 \cdot 1 \cdot 3=16-12=4 > 0\\\\ x_1=\frac{-(-4)-\sqrt{4} }{2 \cdot 1} =\frac{4-2}{2} =\frac{2}{2} =1;~x_2=\frac{-(-4)+\sqrt{4} }{2 \cdot 1} =\frac{4+2}{2} =\frac{6}{2} =3\\\\S=\{1;3\}$

$\displaystyle 4.~~120+120 \cdot \frac{5}{100} =120+\frac{600}{100} =120+6=126\\\\ 126+126 \cdot \frac{5}{100} =126+\frac{630}{100} =126+6,3=132,3$

$\displaystyle 5.~~A(1,7),~B(5,1)\\\\ C-mijlocul~segmentului~AB \Rightarrow x_C=\frac{x_A+x_B}{2} =\frac{1+5}{2} =\frac{6}{2} =3\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~y_C=\frac{y_A+y_C}{2}=\frac{7+1}{2} =\frac{8}{2} =4 \\\\ x_C=3,~y_C=4 \Rightarrow C(3,4)\\\\ OC=\sqrt{(x_C-x_O)^2+(y_C-y_O)^2}= \sqrt{(3-0)^2+(4-0)^2} =\\\\=\sqrt{3^2+4^2}=\sqrt{9+16} =\sqrt{25} =5$

$\displaystyle 6.~~AB=AC=5,~m(\nless B)=45^{\circ}\\\\ AB\equiv AC=5;~m(\nless B)\equiv m(\nless C)=45^{\circ}\\\\ A=\frac{5 \cdot 5}{2} =\frac{25}{2} =12,5$

$SUBIECTUL~II\\\\ 1.~~A=\left(\begin{array}{ccc}1&3\\3&3\end{array}\right),~I_2=\left(\begin{array}{ccc}1&0\\0&1\end{array}\right)\\\\\\ a)~detA=\left|\begin{array}{ccc}1&3\\3&3\end{array}\right|=1 \cdot 3-3 \cdot 3=3-9=-6$

$b)~det(A-xI_2)=-1\\\\ xI_2=x\cdot \left(\begin{array}{ccc}1&0\\0&1\end{array}\right)=\left(\begin{array}{ccc}x&0\\0&x\end{array}\right)\\\\ A-xI_2=\left(\begin{array}{ccc}1&3\\3&3\end{array}\right)-\left(\begin{array}{ccc}x&0\\0&x\end{array}\right)=\left(\begin{array}{ccc}1-x&3\\3&3-x\end{array}\right)\\\\ det(A-xI_2)=\left|\begin{array}{ccc}1-x&3\\3&3-x\end{array}\right|=(1-x)(3-x)-3 \cdot 3=\\\\=3-x-3x+x^2-9=x^2-4x-6$

$\displaystyle det(A-xI_2)=-1 \Rightarrow x^2-4x-6=-1 \Rightarrow x^2-4x-6+1=0 \Rightarrow \\\\ \Rightarrow x^2-4x-5=0\\\\ \Delta=(-4)^2-4 \cdot 1 \cdot (-5)=16+20=36 > 0\\\\ x_1=\frac{-(-4)-\sqrt{36} }{2\cdot 1}=\frac{4-6}{2} =-\frac{2}{2} =-1 \\\\ x_2=\frac{-(-4)+\sqrt{36} }{2\cdot 1}=\frac{4+6}{2} =\frac{10}{2} =5$

$c)~AX=XA,~X=\left(\begin{array}{ccc}2&1\\a&b\end{array}\right)\\\\ \underbrace{\left(\begin{array}{ccc}1&3\\3&3\end{array}\right) \left(\begin{array}{ccc}2&1\\a&b\end{array}\right)}=\left(\begin{array}{ccc}1 \cdot 2+3 \cdot a&1 \cdot 1+3 \cdot b\\3 \cdot 2+3 \cdot a&3 \cdot 1+3 \cdot b\end{array}\right)=\left(\begin{array}{ccc}2+3a&1+3b\\6+3a&3+3b\end{array}\right)\\~~~~~~~~~~~~A X$

$\underbrace{\left(\begin{array}{ccc}2&1\\a&b\end{array}\right) \left(\begin{array}{ccc}1&3\\3&3\end{array}\right)}=\left(\begin{array}{ccc}2 \cdot 1+1 \cdot 3&2 \cdot3+1 \cdot 3\\a \cdot 1+b \cdot 3&a \cdot 3+b \cdot 3\end{array}\right)=\left(\begin{array}{ccc}5&9\\a+3b&3a+3b\end{array}\right)\\~~~~~~~~~~~~XA$

$\displaystyle AX=XA \Rightarrow \left(\begin{array}{ccc}2+3a&1+3b\\6+3a&3+3b\end{array}\right)=\left(\begin{array}{ccc}5&9\\a+3b&3a+3b\end{array}\right) \Rightarrow$

$\Rightarrow \left\{\begin{array}{ccc}2+3a=5 \Rightarrow 3a=5-2 \Rightarrow 3a=3 \Rightarrow a=\cfrac{3}{3}\Rightarrow a=1\\1+3b=9 \Rightarrow 3b=9-1\Rightarrow 3b=8\Rightarrow b=\cfrac{8}{3} \\ 6+3a=a+3b \Rightarrow 6+3 \cdot 1=1+3 \cdot \cfrac{8}{3}\Rightarrow 9=9\\3+3b=3a+3b \Rightarrow 3+3 \cdot \cfrac{8}{3} =3\cdot 1+3\cdot \cfrac{8}{3} \Rightarrow 11=11\end{array}\right\\\\ a=1,~b=\cfrac{8}{3}$

$\displaystyle 2.~~x*y=xy-x-y+2\\\\\\ a)~x*y=(x-1)(y-1)+1\\\\ x*y=xy-x-y+2=xy-x-y+1+1=x(y-1)-(y-1)+1=\\\\=(x-1)(y-1)+1$

$\displaystyle b)~(x*y)*z=x*(y*z)\\\\ (x*y)*z=\underbrace{((x-1)(y-1)+1)}*z=a*z=(a-1)(z-1)+1=\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a\\=((x-1)(y-1)+1-1)(z-1)+1=(x-1)(y-1)(z-1)+1\\\\ x*(y*z)=x*\underbrace{((y-1)(z-1)+1)}=x*b=(x-1)(b-1)+1=\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~b\\=(x-1)((y-1)(z-1)+1-1))+1=(x-1)(y-1)(z-1)+1\\\\ (x*y)*z=x*(y*z) \Rightarrow Legea~"*"~este~asociativa~ \forall~x,y,z \in\mathbb{R}.$

$c)~n*n*n=n\\\\ n*n=(n-1)(n-1)+1=n^2-n-n+1+1=n^2-2n+2\\\\ n*n*n=\Big(n^2-2n+2-1\Big)\big(n-1\big)+1=\Big(n^2-2n+1\Big)\big(n-1\big)+1=\\\\=n^3-n^2-2n^2+2n+n-1+1=n^3-3n^2+3n\\\\ n*n*n=n \Rightarrow n^3-3n^2+3n=n \Rightarrow n^3-3n^2+3n-n =0 \Rightarrow \\\\ \Rightarrow n^3-3n^2+2n=0\Rightarrow n\Big(n^2-3n+2\Big)=0 \Rightarrow n\big(n-1\big)\big(n-2\big)=0\\\\n_2=1\not=0;~n_3=2\not=0$