Question

va rog ajutorrrr.....

Answer 1

[tex]\displaystyle Subiectul~1 \\ \\ 1). \sqrt{2} ,~ \sqrt[3]{4} ,~ \sqrt[4]{5} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \sqrt[n]{x} = \sqrt[ny]{x^y}\\ \\ \sqrt{2} = \sqrt[12]{2^6} = \sqrt[12]{64} ;~~~~~~~~ \sqrt[3]{4} = \sqrt[12]{4^4} = \sqrt[12]{256} ; \\ \\~~~~~~~~~~ \sqrt[4]{5} = \sqrt[12]{5^3} = \sqrt[12]{125} \\ \\ \sqrt[12]{64} \ \textless \ \sqrt[12]{125} \ \textless \ \sqrt[12]{256} \Rightarrow \sqrt{2} \ \textless \ \sqrt[4]{5} \ \textless \ \sqrt[3]{4} [/tex]

[tex]\displaystyle 2).f:R \rightarrow R,~f(x)=4x^2-8x+1 \\ \\ a=4,~b=-8,~c=1 \Rightarrow a\ \textgreater \ 0 \\ \\ f_{min}=- \frac{\Delta}{4a} \\ \\ \Delta=b^2-4ac=(-8)^2-4 \cdot 4 \cdot 1=64-16=48 \\ \\ f_{min}=- \frac{48}{4 \cdot 4} = -\frac{48}{16} =-3 [/tex]

[tex]\displaystyle 3).lg(x-1)+lg(6x-5)=2 \\ \\ \left \{ {{x-1\ \textgreater \ 0} \atop {6x-5\ \textgreater \ 0}} \right. \Rightarrow \left \{ {{x\ \textgreater \ 1~~~~~~~~~~~~~x \in(1,\infty)} \atop {x\ \textgreater \ \frac{5}{6} ~~~~~~~~~~~~x \in \left( \frac{5}{6}, \infty \right) }} \right. \\ \\ \left\begin{array}{ccc}&x \in(1,\infty)\\&x \in \left( \frac{5}{6} ,\infty\right) \\\end{array}\right\} \Rightarrow x \in (1,\infty)=D \\ \\ lg(x-1)+lg(6x-5)=2 \\ \\ lg(x-1)(6x-5)=lg100 \\ \\ (x-1)(6x-5)=100 \\ \\ 6x^2-5x-6x+5=100 \\ \\ 6x^2-11x+5-100=0 [/tex]

[tex]\displaystyle 6x^2-11x-95=0 \\ \\ a=6,~b=-11,~c=-95 \\ \\ \Delta=b^2-4ac=(-11)^2-4 \cdot 6 \cdot (-95)=121+2280=2401 \\ \\ x_1= \frac{11+ \sqrt{2401} }{2 \cdot 6} = \frac{11+49}{12} = \frac{60}{12} =5 \in D \\ \\ x_2= \frac{11- \sqrt{2401} }{2 \cdot 6} = \frac{11-49}{12} =- \frac{38}{12} =- \frac{19}{6} \not \in D \\ \\ S=\{5\}[/tex]

$\displaystyle 4).P= \frac{6}{90} = \frac{1}{15} =0,0(6)$

[tex]\displaystyle 5).A(6,4)~~~~~~~~~~~~~~~~~~~d:2x-3y+1=0 \\ \\ d'-dreapta~cautata,~m_{d'}-panta~ acestei ~drepte \\ \\ m_d-panta~dreptei~d \Rightarrow m_d=- \frac{a}{b} =- \frac{2}{-3} = \frac{2}{3 } \Rightarrow \\ \\ \Rightarrow m_{d'}=- \frac{1}{m_d} =- \frac{1}{ \frac{2}{3} } =1 \cdot \left(- \frac{3}{2} \right)=- \frac{3}{2} \\ \\ Ecuatia ~dreptei~d':y-y_0=m_{d'}(x-x_0) \Rightarrow y-4=- \frac{3}{2} \left(x-6 \right) \Rightarrow [/tex]
[tex]\displaystyle \Rightarrow y-4=- \frac{3}{2} x+ \frac{18}{2} \Rightarrow y-4=- \frac{3}{2} x+9 \Rightarrow 2y-8=-3x+18 \Rightarrow \\ \\ \Rightarrow 3x+2y-18-8=0 \Rightarrow 3x+2y-26=0[/tex]
[tex]\displaystyle 6).sin~a= \frac{1}{3} ,~cos~2a=? \\ \\ cos^2a+sin^2a=1 \Rightarrow cos^2a=1-sin^2a \Rightarrow cos~a=- \sqrt{1-sin^2a} \Rightarrow \\ \\ cos~a=- \sqrt{1- \left(\frac{1}{3} \right)^2} \Rightarrow cos~a=- \sqrt{1- \frac{1}{9} } \Rightarrow cos~a=- \sqrt{ \frac{8}{9} } \Rightarrow \\ \\ \Rightarrow cos~a=- \frac{2 \sqrt{2} }{3} \\ \\ cos~2a=cos(a+a)=cos~a \cdot cos~a-sin~a \cdot sin~a=cos^2a-sin^2a= [/tex]

[tex]\displaystyle =\left(- \frac{2 \sqrt{2} }{3} \right)^2-\left( \frac{1}{3} \right)^2= \frac{8}{9} - \frac{1}{9} = \frac{7}{9} \Rightarrow cos~2a= \frac{7}{9} [/tex]

[tex]\displaystyle Subiectul~2 \\ \\ 1).A= \left(\begin{array}{ccc}0&1&1\\1&0&1\\1&1&0\end{array}\right) \\ \\ a).A^2-A=2I_3[/tex]
[tex]\displaystyle A^2=\left(\begin{array}{ccc}0&1&1\\1&0&1\\1&1&0\end{array}\right) \cdot \left(\begin{array}{ccc}0&1&1\\1&0&1\\1&1&0\end{array}\right)= [/tex]
[tex]=\left(\begin{array}{ccc}0 \cdot 0+1 \cdot 1+1 \cdot 1&0 \cdot 1+1 \cdot 0+1 \cdot 1&0 \cdot 1+1 \cdot 1+1 \cdot 0\\1 \cdot 0+0 \cdot 1+1 \cdot 1&1 \cdot 1+0 \cdot 0+1 \cdot 1&1 \cdot 1+0 \cdot 1 +1 \cdot 0\\1 \cdot 0+1 \cdot 1+0 \cdot 1&1 \cdot 1+1 \cdot 0+0 \cdot 1&1 \cdot 1+1 \cdot 1+0 \cdot 0\end{array}\right) = \\ \\ = \left(\begin{array}{ccc}2&1&1\\1&2&1\\1&1&2\end{array}\right)[/tex]
[tex]\displaystyle A^2-A=\left(\begin{array}{ccc}2&1&1\\1&2&1\\1&1&2\end{array}\right)- \left(\begin{array}{ccc}0&1&1\\1&0&1\\1&1&0\end{array}\right)= \\ \\ = \left(\begin{array}{ccc}2-0&1-1&1-1\\1-1&2-0&1-1\\1-1&1-1&2-0\end{array}\right)= \left(\begin{array}{ccc}2&0&0\\0&2&0\\0&0&2\end{array}\right) =2 \cdot \left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right) =2I_3 \Rightarrow \\ \\ \Rightarrow A^2-A=2I_3[/tex]