35 A => on the flip side

36 A => in

37 C => have been derived 

38 D => tremediously

39 B => becomes 

40 D => of which

41 C => minority languages

42 C => lie under 

43 B => happens 

44 B => percent

45 C => have been born

46 B=> of whom

47 C => as well as 

48 A bỏ đi

49 C => ethnicity

50 C => sacrificing

3) \(\sqrt{\left(x-2\right)\left(x+1\right)}\) thì (x-2)(x+1)>0

=> x² -x-2>0

=> x² - x - \(\dfrac{1}{2}\)- \(\dfrac{3}{2}\)>0

= (x+\(\dfrac{1}{4}\))² - 3/2 >0

=> x+ 1/4>3/2

=> x>5/4

4) Có x đâu mà tìm bạn??

4) \(\sqrt{x^2+2x+1}\) + \(\sqrt{x^2-2x+1}\)= \(\sqrt{\left(x+1\right)^2}\) + \(\sqrt{\left(x-1\right)^2}\) 

=> /x+1/+/x-1/ = 2

=> /2x/ = 2

=> 2x=2

=> x=1

Để biểu thức có nghĩa thì (x-2)(x+1)>0

\(\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -1\end{matrix}\right.\)

\(a^3+b^3=\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}-\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right)}\)

\(=\sqrt{6}-\sqrt{2}-\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{4}=0\)

\(\Rightarrow a=-b\Rightarrow a^5+b^5=0\)

a)\(\left\{{}\begin{matrix}\xi_b=\xi_1+\xi_2=6+3=9V\\r_b=r_1+r_2=2+1=3\Omega\end{matrix}\right.\)

b)CTM ngoài: \(R_1nt\left(R_2//R_3\right)\)

\(R_{23}=\dfrac{R_2\cdot R_3}{R_2+R_3}=\dfrac{2\cdot8}{2+8}=1,6\Omega\)

\(R_N=R_1+R_{23}=4,4+1,6=6\Omega\)

c)\(I_1=I_{23}=I=\dfrac{\xi_b}{r_b+R_N}=\dfrac{9}{3+6}=1A\)

\(U_1=I_1\cdot R_1=1\cdot4,4=4,4V\)

\(U_2=U_3=U_{23}=I_{23}\cdot R_{23}=1\cdot1,6=1,6V\)