1.A
2.A
3.B
4.C
5.B
6.C
7.A
8.A
9.B
10.A
11.B
12.A
13.C
14.B
15.B
16.A
17.A
18.A
19.A
20.C

bài j vậy bn

? đề đâu

chiu ma lai

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

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

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

Bài 1:

\(a,\Rightarrow-30x=12\cdot25=300\Rightarrow x=-10\\ b,\Rightarrow24x=\dfrac{7}{5}\cdot12=\dfrac{84}{5}\Rightarrow x=\dfrac{7}{10}\\ c,\Rightarrow\dfrac{4}{3}\cdot\dfrac{5}{4}=\dfrac{2}{3}:\dfrac{1}{10}x\\ \Rightarrow\dfrac{5}{3}=\dfrac{20}{3}x\Rightarrow x=\dfrac{1}{4}\\ d,\Rightarrow-2,5x=-40\Rightarrow x=16\\ e,\Rightarrow x^2=100\Rightarrow\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)

A

A