\(x^2-6y^2=1\)

Chọn C

ảnh lỗi

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

e: \(E=\dfrac{x^2-9-x^2+4-x^2+9}{\left(x+3\right)\left(x-2\right)}\)

\(=\dfrac{x+2}{x+3}\)

a: \(A=\dfrac{4x^2+x^2-2x+1+x^2+2x+1}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{6x^2+2}{\left(x-1\right)\left(x+1\right)}\)

\(A=\dfrac{-4x^2+x^2-2x+1-x^2-2x-1}{\left(1-x\right)\left(1+x\right)}=\dfrac{-4x\left(x+1\right)}{\left(1-x\right)\left(1+x\right)}=\dfrac{4x}{x-1}\\ C=\dfrac{-x^2-4x-4+x^2-4x+4-4x^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-4x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4x}{2-x}\\ E=\dfrac{x^2-9-x^2+4x-4-x^2+9}{\left(x-2\right)\left(x+3\right)}=\dfrac{-\left(x-2\right)^2}{\left(x-2\right)\left(x+3\right)}=\dfrac{2-x}{x+3}\)

1 are

2 am

3 is

4 are

5 are

6 are

7 is

8 is

9 is

10 are

IV

1 is writing

2 are losing

3 is having

4 is staying

5 am not lying

6 is always using

7 are having

8 Are you playing

9 are not touching

10 Is - listening

11 Is- winning

12 am not staying

13 is not working

14 is not reading

15 isn't raining

16 am not listening

17 Are they making

18 Are you doing

19 Is - sitting

20 is - doing

21 are-putting

22 are-wearing

23 is-studying

2, am

3, is

4,are

5,are

6,are

7,is

8,is

9,is

10,are

IV

1,2,7 OK

3,is having

4,has stayed

5,am not lying

6,always uses

8,Are-playing

9,not to touch

10,Is-listening

11,Are-winning

12,am not staying

13,isn't working

14,isn't reading

15,isn't raining

16,am not listening

17,Are-making

18,Are-doing

19,Is-sitting

20,is-doing

21,do-putting

22,do-wear

23,is-studying

5: \(=\dfrac{1}{2}\cdot10-\dfrac{1}{2}=\dfrac{1}{2}\cdot9=\dfrac{9}{2}\)

a, Thay tọa độ điểm ( 2;5 ) vào hàm số ta được ;

\(2\left(2m-1\right)+m-3=5\)

\(\Rightarrow m=2\)

b, - Gọi điểm cố định hàm số đi qua là M (x0; y0 ) ta được :

\(\left(2m-1\right)x_0+m-3=y_0\)

\(\Leftrightarrow2mx_0-x_0+m-3-y_0=0\)

\(\Leftrightarrow m\left(2x_0+1\right)-x_0-y_0-3=0\)

- Để hàm số luôn đi qua điểm cố định \(\Leftrightarrow\left\{{}\begin{matrix}2x_0+1=0\\x_0+y_0+3=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{5}{2}\end{matrix}\right.\)

Vậy điểm cố định mà hàm số đi qua là : M ( -1/2; -5/2 )

c, - Thay điểm có hoành độ là \(\sqrt{2}-1\) vào hàm số ta được :

\(\left(\sqrt{2}-1\right)\left(2m-1\right)+m-3=0\)

\(\Leftrightarrow m=\dfrac{6+5\sqrt{2}}{7}\)

Vậy ...