2 2 2 2X X X X 2X X M 2      0,251 2 1 2 1 23 3 3 3(1,0X...

Bài 2

2

2

2

2

x

x

x

x

2x x

m

2

 

0,25

1

2

1

2

1 2

3

3

3

3

(1,0

x

x

x

x

3x x x

x

m

3m

   

1

2

1

2

1 2

1

2

điểm)

Do đó:

       

2

2

3

3

2

3

x

x

x

x

m

2 m

3m

1

2

1

2

5

5

2

3

3

2

5

3

3

x

x

x x

x x

m

2m

3m

6m

1

2

1

2

1

2

5

5

2

2

5

3

x

x

x x

x

x

m

5m

6m

5

5

5

3

x

x

m m

5m

6m

1

2

x

x

m

5m

5m

Vậy:

x

1

5

x

2

5

Z

vì m

Z

.

0,25