1. B

2. C

Vì \(\text{∠}zDx=90\text{ °}\) nên \(Dx\perp Dz\)

Ta có \(\left\{{}\begin{matrix}Dx\perp Dz\\Ey\perp Dz\end{matrix}\right.\) \(\Rightarrow Dx//Ey\)

a. Vì \(R_1ntR_2\) nên \(R_{12}=R_1+R_2=15+25=40\left(\text{Ω}\right)\)

Vì \(R_{12}//R_3\) nên \(\dfrac{1}{R_{td}}=\dfrac{1}{R_{12}}+\dfrac{1}{R_3}\Rightarrow R_{td}=\dfrac{R_{12}.R_3}{R_{12}+R_3}=\dfrac{40.10}{40+10}=8\left(\text{Ω}\right)\)

b. Ta có \(I=\dfrac{U}{R_{td}}=\dfrac{12}{8}=1,5\left(A\right)\)

mà \(U_{12}=U_3\Leftrightarrow R_{12}.I_{12}=R_3.I_3\Leftrightarrow40I_{12}=10I_3\Leftrightarrow I_3=4I_{12}\) (1)

mặt khác, ta có \(I=I_{12}+I_3\) (2)

Từ (1) và (2) \(\Rightarrow I_{12}+4I_{12}=1,5\Rightarrow I_{12}=0,3\left(A\right)\)

\(\Rightarrow I_3=I-I_{12}=1,5-0,3=1,2\left(A\right)\)

c. Ta có \(R_{td'}=\dfrac{R_{2x}.R_3}{R_{2x}+R_3}=\dfrac{\left(25+R_x\right)10}{R_x+25+10}=\dfrac{250+10R_x}{35+R_x}=7,5\left(\text{Ω}\right)\)

\(\Rightarrow R_x=5\left(\text{Ω}\right)\)

Bạn chụp thêm hình vẽ nữa chứ không biết mắc song song hay nối tiếp để làm 

Ta có \(\left(n+3\right)^2\ge0\forall x\) \(\Rightarrow4-\left(n+3\right)^2\le4\forall x\)

Dấu "=" xảy ra khi \(n+3=0\Leftrightarrow n=-3\)

Vậy \(A_{min}=4\) khi \(x=-3\)

Kate accused Paul of breaking his special pen

a. \(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}-x+\sqrt{x}-1}\right):\left(\dfrac{x+\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}+\dfrac{1}{x+1}\right)\)

\(=\dfrac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}:\dfrac{x+\sqrt{x}+\sqrt{x}+1}{\left(x+1\right)\left(\sqrt{x}+1\right)}\)

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

b. Để \(P< \dfrac{1}{2}\) thì \(\dfrac{\sqrt{x}-3}{\sqrt{x}+1}< \dfrac{1}{2}\) \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{1}{2}< 0\) \(\Leftrightarrow\dfrac{\sqrt{x}-3}{2\left(\sqrt{x}+1\right)}< 0\) 

mà \(2\left(\sqrt{x}+1\right)\ge2>0\forall x\) \(\Rightarrow\sqrt{x}-3< 0\Rightarrow\sqrt{x}< 3\Leftrightarrow x< 9\)

c. Để \(P=\dfrac{1}{3}\) thì\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{1}{3}\Rightarrow3\left(\sqrt{x}-1\right)=\sqrt{x}+1\Rightarrow2\sqrt{x}=4\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

d. Ta có \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)

Để \(P\) nhận giá trị nguyên thì \(2⋮\left(\sqrt{x}+1\right)\Rightarrow\left(\sqrt{x}+1\right)\inƯ\left(2\right)\)

\(\Rightarrow\sqrt{x}+1\in\left\{1;-1;2;-2\right\}\) \(\Rightarrow x\in\left\{0;1\right\}\)

e. Ta có \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)

Với mọi \(x\ge0\) thì \(\sqrt{x}\ge0\) \(\Rightarrow\sqrt{x}+1\ge1\Rightarrow\dfrac{2}{\sqrt{x}+1}\le2\Rightarrow P=1-\dfrac{2}{\sqrt{x}+1}\ge-1\) 

Dấu "=" xảy ra khi \(\sqrt{x}+1=1\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)

Vậy \(P_{min}=-1\) khi \(x=0\)

4. Even though Tom worked very hard, he failed the exam

5. Even though Linda got the highest result in the class, she still had problems with the teacher

6. Even though he is a little overweight, he is actually quite fit

7. In spite of having a good salary, he was unhappy in his job

8. Despite earning a low salary, she gave money to her parents

9. Despite her physical handicap, she has become a successful woman

10. Although he was tired, he walked to the station

D

1. The film on television made us so boring that we went  to bed early.

   → The film on telvevision made us so bored that we went to bed early

  2. Everyone stopped to laugh when we came.

   → Everyone stopped laughing when we came 

  3. They are going back to the USA for two months.

   → They have gone back to the USA for two months

  4. He warned us to be careful of strangers and not go out at night.

   → He warned us to be careful of strangers and not to go out at night

  5. I have been looking for this book for months and in the end I have found them.

   → I have been looking for this book for months and in the end I have founnd it