a. \(\sqrt{-2x+3}\)       

ĐKXĐ: x < 0

b. \(\sqrt{\dfrac{2}{x^2}}\)

ĐKXĐ: x \(\ne\) 0

c. \(\sqrt{\dfrac{4}{x+3}}\)

ĐKXĐ: x > -3

d. \(\sqrt{\dfrac{-5}{x^2+6}}\)

ĐKXĐ: x vô nghiệm 

4. a. x² - 7

= x² - \(\left(\sqrt{7}\right)^2\)

= \(\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)\)

b. x² - \(2\sqrt{2}x\) + 2

= x² - \(2\sqrt{2}x\) + \(\left(\sqrt{2}\right)^2\)

= (x - \(\sqrt{2}\))²

c. x² + \(2\sqrt{13}x\) + 13

= x² + \(2\sqrt{13}x\) + \(\left(\sqrt{13}\right)^2\)

= \(\left(x+\sqrt{13}\right)^2\)

.

Section B

9 A

10 C

11 A

12 D

13 B

14 D

15 B

16 A

17 C

18 D

19 D

20 C

21 C

22 B

23 C

24 C

25 C

26 B

27 A

28 D

29 A

30 C

31 A

32 A

33 Unless she keeps her feet dry, she willl catch a cold

34 THe last time I saw Tom was 2 weeks ago

35 It's interesting to spend our weekend in the countryside

36 I spent 4 hours reading the first chapter of the book

nhiều thế

a: n=4

b: n=2

lỗi r ạ

?

đề:)?

Bài 3:

1: ĐKXĐ: x<>-y/2

\(\left\{{}\begin{matrix}\dfrac{4x-y}{2x+y}-\dfrac{x-2y}{2x+y}=2\\3x-y=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{4x-y-x+2y}{2x+y}=2\\3x-y=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x-y=2\left(2x+y\right)\\3x-y=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x-y-4x-2y=0\\3x-y=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-x-3y=0\\3x-y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x-9y=0\\3x-y=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-10y=4\\3x-y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{2}{5}\\3x=y+4=-\dfrac{2}{5}+4=\dfrac{18}{5}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{2}{5}\\x=\dfrac{6}{5}\end{matrix}\right.\left(nhận\right)\)

2: 

a: Thay x=1 và y=2 vào (d_m), ta được:

\(1\left(2m-1\right)-m+1=2\)

=>2m-1-m+1=2

=>m=2

Thay m=2 vào (d_m), ta được:

\(y=\left(2\cdot2-1\right)x-2+1=3x-1\)

Vẽ đồ thị:

b: tọa độ giao điểm của (d_m) với trục hoành là:

\(\left\{{}\begin{matrix}y=0\\\left(2m-1\right)x-m+1=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=0\\x\left(2m-1\right)=m-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=\dfrac{m-1}{2m-1}\end{matrix}\right.\)

Để x nguyên thì \(m-1⋮2m-1\)

=>\(2m-2⋮2m-1\)

=>\(2m-1-1⋮2m-1\)

=>\(-1⋮2m-1\)

=>\(2m-1\in\left\{1;-1\right\}\)

=>\(2m\in\left\{2;0\right\}\)

=>\(m\in\left\{1;0\right\}\)