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
ớ
x≥0;
x≠4
)
√
x
−2
+
b)
B=
x
−10
Gi i:
ả
√
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
)
=