1√3=4 √X−2√X−1X−√X−2−√X+1 (V I Ớ X≥0; X≠4 ) √X−2+B) B= X−10GI I

1

.

1

3

=

4

3.

1

3

=

4

x

−2

x

−1

x

x

−2

x

+1

(v i

x≥0;

x≠4

)

x

−2

+

b)

B=

x

−10

Gi i:

x−1

x−2

x−

x−2

x−2

+

x

+1

Ta có

B=

x−10

=

x−10

(

x−2

) (

x+1

)

=

x−10−

(

x−1

) (

x

+1

)

+

(

x−2

) (

x−2

)

(

x−2

) (

x+1

)

x−10−

x+

1+

x

−4

x

+4

=

x−10−(

x−1)+

(

x−

4

x+4

)

(

x−2

) (

x+1

)

=

(

x−2

) (

x+1

)

(

x

+1

)(

x−5

)

x−5

=

x−

4

x−5

(

x−2

) (

x+1

)

=