Question

Am nevoie de ajutor !

Answer 1

$\displaystyle \mathtt{2).~~~f:\mathbb{R}\rightarrow\mathbb{R},~f(x)=x+1}\\\\ \mathtt{g:\mathbb{R}\rightarrow\mathbb{R},~g(x)=x^2-x+2}\\ \\ \mathtt{f(x)=g(x)}\\\\\mathtt{x+1=x^2-x+2}\\\\\mathtt{-x^2+x+x+1-2=0}\\\\ \mathtt{-x^2+2x-1=0\Big|\cdot(-1)}\\\\\mathtt{x^2-2x+1=0}\\\\ \mathtt{a=1,~b=-2,~c=1}\\\\\mathtt{\Delta=b^2-4ac=(-2)^2-4\cdot 1\cdot 1=4-4=0}\\ \\ \mathtt{x_1=x_2=-\frac{b}{2a}=-\frac{-2}{2\cdot1}= \frac{2}{2}=1 }$

$\displaystyle \mathtt{3).~~~3^{x^2+3}=3 \cdot 3^{3x}}\\ \\ \mathtt{3^{x^2+3}=3^{1+3x}}\\ \\ \mathtt{x^2+3=1+3x}\\ \\ \mathtt{x^2-3x+3-1=0}\\ \\ \mathtt{x^2-3x+2=0}\\ \\ \mathtt{a=1,~b=-3,~c=2}\\ \\ \mathtt{\Delta=b^2-4ac=(-3)^2-4\cdot1\cdot2=9-8=1\ \textgreater \ 0}\\ \\ \mathtt{x_1= \frac{-(-3)-1}{2 \cdot 1}= \frac{3-1}{2} = \frac{2}{2} =1 }\\ \\ \mathtt{x_2= \frac{-(-3)+1}{2}= \frac{3+1}{2}= \frac{4}{2}=2 }$

$\displaystyle \mathtt{4).~~~P= \frac{nr.~cazuri~favorabile}{nr.~cazuri~posibile} }\\ \\ \mathtt{Numerele~naturale~de~dou\u{a}~cifre:10,11,12,...,99\Rightarrow 90~ cazuri}\\ \mathtt{posibile}\\ \\ \mathtt{Numerele~divizibile~cu~3~\c{s}i~5:15,30,45,60,75,90\Rightarrow 6~cazuri}\\ \mathtt{favorabile}\\ \\ \mathtt{P= \frac{6}{90}= \frac{1}{15} }$

$\displaystyle \mathtt{5).~~~A(0,2),~B(2,4),~C(m,0)}\\ \\ \mathtt{Punctele~A(0,2),~B(2,4),~C(m,0)~sunt~coliniare\Leftrightarrow \left|\begin{array}{ccc}\mathtt0&\mathtt2&\mathtt1\\\mathtt2&\mathtt4&\mathtt1\\\mathtt m&\mathtt0&\mathtt1\end{array}\right|=0}\\\\\mathtt{0\cdot 4 \cdot 1+1 \cdot 2 \cdot 0+2 \cdot 1 \cdot m-1 \cdot 4 \cdot m-2\cdot 2 \cdot 1-0 \cdot 1 \cdot 0=0}\\ \\ \mathtt{0+0+2m-4m-4-0=0}\\\\\mathtt{-2m-4=0}\\\\ \mathtt{-2m=0+4}\\\\\mathtt{-2m=4}\\\\\mathtt{m=-\frac{4}{2} }\\\\\mathtt{m=-2}$

$\displaystyle \mathtt{6).~~~AB=4,~AC=8,~A= \frac{\pi}{3}}\\ \\ \mathtt{a^2=b^2+c^2-2 \cdot b \cdot c\cdot cos~A}\\ \\ \mathtt{BC^2=AC^2+AB^2-2 \cdot AC \cdot AB \cdot cos \frac{\pi}{3} }\\ \\ \mathtt{BC^2=8^2+4^2-2 \cdot 8\cdot4\cdot cos60^\circ}\\ \\ \mathtt{BC^2=64+16-64 \cdot \frac{1}{2} }\\ \\ \mathtt{BC^2=80-32}\\ \\ \mathtt{BC^2=48}\\ \\ \mathtt{BC= \sqrt{48}}\\ \\ \mathtt{BC=4 \sqrt{3} }$