Question

Va rog la ce exerciții știți !!

Answer 1

$\displaystyle \mathtt{Subiectul ~1}\\ \\ \mathtt{1.~~~\left( \frac{5}{3}\right)^{-1}- \sqrt[\mathtt3]{\mathtt{125}}+log_ {\sqrt[\mathtt3]{\mathtt2}}8= \frac{3}{5}- \sqrt[\mathtt3]{\mathtt{5^3}} +log_{2^{ \frac{1}{3} }}8=}\\ \\ \mathtt{= \frac{3}{5} -5+ \frac{1}{ \frac{1}{3} }log_28= \frac{3}{5}-5+3log_22^3= \frac{3}{5}-5+3\cdot3log_22=}\\ \\ \mathtt{= \frac{3}{5}-5+9= \frac{3-25+45}{5}= \frac{23}{5}}$

$\displaystyle \mathtt{2.~~~log_2\left(x^2+4\right)=2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~C.E.~x^2+4\ \textgreater \ 0}\\ \\ \mathtt{log_2\left(x^2+4\right)=log_24}\\ \\ \mathtt{x^2+4=4\Rightarrow x=0}\\ \\ \mathtt{0^2+4\ \textgreater \ 0\Rightarrow 4\ \textgreater \ 0~A\Rightarrow x=0~este~solu\c{t}ie~a~ecua\c{t}iei~~~~~~~S=\{0\}}}$

$\displaystyle \mathtt{3.~~~C_1^0+C_2^1+C_3^2+C_4^3+C_5^4=}\\ \\ \mathtt{=1+2+ \frac{3!}{(3-2)! \cdot 2!}+ \frac{4!}{(4-3)!\cdot 3!}+ \frac{5!}{(5-4)! \cdot 4!} =}}\\ \\ \mathtt{=1+2+3+4+5=15}$

$\displaystyle \mathtt{4.~~~f,g:\mathbb{R}\rightarrow\mathbb{R},~f(x)=2x-1,~g(x)=6-x}\\ \\ \mathtt{f(x)=g(x)\Rightarrow 2x-1=6-x\Rightarrow 2x+x=6+1\Rightarrow 3x=7\Rightarrow x= \frac{7}{3} }\\ \\ \mathtt{f\left( \frac{7}{3} \right)=6- \frac{7}{3}= \frac{18-7}{3}= \frac{11}{3} \Rightarrow f\left( \frac{7}{3} \right)= \frac{11}{3} }\\ \\ \mathtt{M\left( \frac{7}{3} ,~ \frac{11}{3} \right)}$

$\displaystyle \mathtt{5.~~~A= \frac{3l^2 \sqrt{3} }{2}}\\ \\ \mathtt{P=6l\Rightarrow 6l=6\Rightarrow l= \frac{6}{6}\Rightarrow l=1 }\\ \\ \mathtt{A= \frac{3\cdot1^2 \sqrt{3} }{2} = \frac{3 \sqrt{3} }{2} }$

$\displaystyle \mathtt{6.~~~A(1,2),~B(-5,-4)}\\ \\ \mathtt{AB= \sqrt{(x_A-x_B)^2+(y_A-y_B)^2} }\\ \\ \mathtt{AB= \sqrt{((1-(-5))^2+(2-(-4))^2} = \sqrt{(1+5)^2+(2+4)^2}= }\\ \\ \mathtt{= \sqrt{6^2+8^2}= \sqrt{36+64}= \sqrt{100} =10}\\ \\ \mathtt{AB=10}$

$\displaystyle \mathtt{Subiectul~2}\\ \\ \mathtt{1.~~~ A= \left(\begin{array}{ccc}\mathtt1&\mathtt{ \sqrt{2}} &\mathtt1\\\mathtt0&\mathtt1&\mathtt{ \sqrt{2}} \\\mathtt0&\mathtt0&\mathtt1\end{array}\right)}\\ \\ \mathtt{A^2=?~~~~A^3=?}$

$\displaystyle \mathtt{A^2=A\cdot A=\left(\begin{array}{ccc}\mathtt1&\mathtt{ \sqrt{2}} &\mathtt1\\\mathtt0&\mathtt1&\mathtt{ \sqrt{2}} \\\mathtt0&\mathtt0&\mathtt1\end{array}\right)\cdot\left(\begin{array}{ccc}\mathtt1&\mathtt{ \sqrt{2}} &\mathtt1\\\mathtt0&\mathtt1&\mathtt{ \sqrt{2}} \\\mathtt0&\mathtt0&\mathtt1\end{array}\right)= }$

$\displaystyle \mathtt{=\left(\begin{array}{ccc}\mathtt{1\cdot1+ \sqrt{2}\cdot0+1\cdot0 }&\mathtt{ 1\cdot \sqrt{2}+ \sqrt{2}\cdot1+1\cdot0 } &\mathtt{1\cdot1+ \sqrt{2}\cdot \sqrt{2}+ 1\cdot1}\\\mathtt{0\cdot1+1\cdot0+ \sqrt{2}\cdot0 }&\mathtt{0\cdot \sqrt{2}+1\cdot1+ \sqrt{2}\cdot0 }&\mathtt{0\cdot1+1\cdot \sqrt{2}+\sqrt{2}\cdot1 } \\\mathtt{0\cdot1+0\cdot0+1\cdot0}&\mathtt{0\cdot \sqrt{2}+0\cdot1+1\cdot0 }&\mathtt{0\cdot1+0\cdot \sqrt{2}+1\cdot1 }\end{array}\right)=}$

$\mathtt{=\left(\begin{array}{ccc}\mathtt{1+0+0}&\mathtt{ \sqrt{2}+ \sqrt{2}+0 } &\mathtt{1+2+1}\\\mathtt{0+0+0}&\mathtt{0+1+0}&\mathtt{0+ \sqrt{2}+ \sqrt{2} } \\\mathtt{0+0+0}&\mathtt{0+0+0}&\mathtt{0+0+1}\end{array}\right)=\left(\begin{array}{ccc}\mathtt1&\mathtt{2 \sqrt{2}} &\mathtt4\\\mathtt0&\mathtt1&\mathtt{ 2\sqrt{2}} \\\mathtt0&\mathtt0&\mathtt1\end{array}\right)}$

$\Displaystyle \mathtt{A^3=A^2\cdot A=\left(\begin{array}{ccc}\mathtt1&\mathtt{2 \sqrt{2}} &\mathtt4\\\mathtt0&\mathtt1&\mathtt{ 2\sqrt{2}} \\\mathtt0&\mathtt0&\mathtt1\end{array}\right)\cdot \left(\begin{array}{ccc}\mathtt1&\mathtt{ \sqrt{2}} &\mathtt1\\\mathtt0&\mathtt1&\mathtt{ \sqrt{2}} \\\mathtt0&\mathtt0&\mathtt1\end{array}\right)=}$

$\displaystyle \mathtt{= \left(\begin{array}{ccc}\mathtt{1+0+0}&\mathtt{ \sqrt{2}+2 \sqrt{2}+0 } &\mathtt{1+4+4}\\\mathtt{0+0+0}&\mathtt{0+1+0}&\mathtt{0+ \sqrt{2}+2 \sqrt{2} } \\\mathtt{0+0+0}&\mathtt{0+0+0}&\mathtt{0+0+1}\end{array}\right)= \left(\begin{array}{ccc}\mathtt1&\mathtt{3\sqrt{2}} &\mathtt9\\\mathtt0&\mathtt1&\mathtt{3\sqrt{2}} \\\mathtt0&\mathtt0&\mathtt1\end{array}\right)}$

$\displaystyle \mathtt{b)~det(A)=? }\\ \\ \mathtt{det(A)=\left|\begin{array}{ccc}\mathtt1&\mathtt{ \sqrt{2}} &\mathtt1\\\mathtt0&\mathtt1&\mathtt{ \sqrt{2}} \\\mathtt0&\mathtt0&\mathtt1\end{array}\right|=1\cdot1\cdot1+1\cdot0\cdot0+ \sqrt{2}\cdot \sqrt{2}\cdot0 -1\cdot1\cdot0- }\\ \\ \mathtt{- \sqrt{2}\cdot0\cdot1-1\cdot \sqrt{2}\cdot0=1 }\\ \\ \mathtt{~~~~~~~~~~~~~~~~~det(A)=1}$