Question

Va rog .. Asemanator subiectului II primesc la teza :((( Si vreau sa inteleg si eu .. va rooog <333 Am nevoie de rezolvarea exerciilor 1,2,3.. 

Answer 1

Subiectul II

[tex]\displaystyle 1a) \\ \left( \sqrt{2+ \sqrt{3} }- \sqrt{2- \sqrt{3} }\right)^2 + m \sqrt{3}=-2 \\ \left( \sqrt{2+ \sqrt{3} }\right)^2 -2 \sqrt{2+ \sqrt{3} }\times \sqrt{2- \sqrt{3} } +\left( \sqrt{2-\sqrt{3} }\right)^2 +m \sqrt{3}=-2\\ 2+ \sqrt{3} -2 \sqrt{(2+ \sqrt{3}) \times (2- \sqrt{3}) } + 2-\sqrt{3} +m \sqrt{3}=-2\\ 2+ \sqrt{3} + 2-\sqrt{3} -2 \sqrt{(2+ \sqrt{3}) \times (2- \sqrt{3}) } +m \sqrt{3}=-2\\ 2+2 - 2\sqrt{4- 3} +m \sqrt{3}=-2 [/tex]

[tex]\displaystyle 4 - 2\sqrt{1} +m \sqrt{3}=-2 \\ 2+m \sqrt{3}=-2 \\ m \sqrt{3}=-2-2 \\m \sqrt{3}=-4 \\ m = \frac{-4}{\sqrt{3}} = \boxed{- \frac{4\sqrt{3}}{3} } [/tex]


[tex]\displaystyle 1b) \\ 9x^2 + 4y^2-12x-4y+5=0 \\ 9x^2 + 4y^2-12x-4y+4+1=0 \\ (9x^2 -12x+4)+( 4y^2-4y+1) =0\\ ((3x)^2 -2\cdot 3x\cdot 2+2^2)+( (2y)^2-2\cdot 2y\cdot 1 + 1^2) =0\\ (3x -2)^2+( 2y-1)^2 =0 ~~\Longrightarrow~~ (3x -2)^2=0~~si~~ ( 2y-1)^2 =0\\ \boxed{x = \frac{2}{3}} ~~~radacina ~dubla \\ \boxed{y= \frac{1}{2} } ~~~radacina ~dubla[/tex]


[tex]1c) ~~~ a\ \textless \ 1 \\ |a-1|-|a^2-1|+|3a-3| = \\ =1-a - (1-a^2) + 3-3a = \\ =1-a - 1+a^2 + 3-3a =\\ =\boxed{a^2 -4a+3} = \\ =a^2 -a-3a+3 = \\ =a(a-1)-3(a-1) = \\ = \boxed{(a-1)(a-3)} [/tex]


[tex]\displaystyle 2a) \\ E(x)=\left( \frac{x-1}{x+2} + \frac{x+1}{2-x}+ \frac{2+7x}{x^2-4} \right)\cdot \left( x+1 +\frac{3-6x}{x+1} \right)= \\ \\ =\left( \frac{x-1}{x+2} - \frac{x+1}{x-2}+ \frac{2+7x}{x^2-4} \right)\cdot \left(\frac{(x+1)^2+ 3-6x}{x+1} \right)= \\ \\ = \frac{(x-1)(x-2)-(x+1)(x+2)+2+7x}{x^2-4} \cdot \frac{x^2 +2x+1+3-6x }{x+1}= \\ \\ =\frac{x^2-3x+2-(x^2+3x+2)+2+7x}{x^2-4} \cdot \frac{x^2 -4x+4}{x+1}= [/tex]


[tex]\displaystyle =\frac{x^2-3x+2-x^2-3x-2+2+7x}{x^2-4} \cdot \frac{(x-2)^2}{x+1}= \\ \\ =\frac{x^2-x^2 +7x -3x -3x +2 -2+2}{x^2-4} \cdot \frac{(x-2)^2}{x+1}= \\ \\ =\frac{x+2}{(x+2)(x-2)} \cdot \frac{(x-2)(x-2)}{x+1}= \boxed{\frac{x-2}{x+1}} [/tex]


[tex]\displaystyle 2b) \\ \frac{x-2}{x+1} = \frac{x+1-3}{x+1}=\frac{x+1}{x+1}-\frac{3}{x+1} \\ \\ D_3 = \{-3;~-1;~1;~3\} \\ x+1=-3 ~~\Longrightarrow~~x_1=-4 \\ x+1=-1 ~~\Longrightarrow~~x_2=-2 \\ x+1= 1 ~~\Longrightarrow~~x_3=0 \\ x+1=3 ~~\Longrightarrow~~x_4=2 [/tex]



[tex]3a) \\ \text{AD si AF sunt doua drepte incluse in planul (DAF)} \\ AB \underline{~|~} AD ~~si~~ AB \underline{~|~} AF \\ \text{Daca o dreapta este perpendiculara pe doua drepte din plan, } \\ \text{atunci dreapta este perpendiculara pe plan} \\ =\ \textgreater \ AB \underline{~|~} (DAF) \\ EF || AB \\ =\ \textgreater \ EF \underline{~|~} (DAF) [/tex]


[tex]3b) \\ EF || AB \\ CD || AB \\ =\ \textgreater \ ~~EF || BC =\ \textgreater \ ~~\text{Dreptele EF si CD sunt coplanare} \\ =\ \textgreater \ ~~\text{Punctele E, F, C, D sunt coplanare} \\ \\ 3c)\\ \text{EF = CD si }EF || CD \\ \text{DF= CE si }DF || CE \\ EF \underline{~|~} FD ~~si~~ EF \underline{~|~} CE\\ =\ \textgreater \ EFCD = dreptunghi \\ EF=CD = AB = 15~cm \\ DF = CE = \sqrt{20^2 + 10^2}= \sqrt{400 + 100}= \sqrt{500}=10 \sqrt{5}~cm \\ A = 15 \times 10 \sqrt{5} = 150 \sqrt{5}~cm^2 ~P=2(15+10 \sqrt{5})cm[/tex]


[tex]3d) \\ \text{Alegem un punct pe AB din care ducem o perpendiculara pe (CDF)} \\ \text{Alegem punctul A, din care perpendiculara pe (CDF), cade pe DF} \\ \text{Aceasta este inaltimea h a } \Delta ADF \text{ dusa din A pe DF} \\ \\ \displaystyle h = \frac{AD \times AF}{DF} = \frac{20 \times 10}{10 \sqrt{5}} = \frac{20\sqrt{5} }{ 5}=4\sqrt{5}~cm[/tex]