II.

1. D

2. A

3. B

4. C

5. D (đoán thế)

III.

1. playing

2. car

3. hobby

4. flu

5. weight

6. homeless

IV:

1. C

2. C

3. D

I:Đáp án

1a

2b

3c

4a

5c

Lời giải:

$\frac{b}{2}=\frac{c}{5}\Leftrightarrow \frac{b}{4}=\frac{c}{10}$

Vậy: $\frac{a}{3}=\frac{b}{4}=\frac{c}{10}$

Áp dụng tính chất dãy tỉ số bằng nhau:

$\frac{a}{3}=\frac{b}{4}=\frac{c}{10}=\frac{a+b-c}{3+4-10}=\frac{3}{-3}=-1$

$\Rightarrow a=-1.3=-3; b=-1.4=-4; c=-1.10=-10$

3: Ta có: A=B|x-4|

\(\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}-5}:\dfrac{1}{\sqrt{x}-5}=\left|x-4\right|\)

\(\Leftrightarrow\left|x-4\right|=\sqrt{x}-2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=\sqrt{x}-2\left(x\ge4;x\ne25\right)\\x-4=2-\sqrt{x}\left(0< x< 4\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{x}-2=0\\x+\sqrt{x}-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=4\left(loại\right)\end{matrix}\right.\)

1: Thay x=9 vào A, ta được:

\(A=\dfrac{3-2}{3-5}=\dfrac{-1}{-2}=\dfrac{1}{2}\)

2: Ta có: \(B=\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{x-25}\)

\(=\dfrac{3\sqrt{x}-15+20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)

\(=\dfrac{1}{\sqrt{x}-5}\)