题目内容
阅读下列材料,并回答问题:∵1 |
1×3 |
1 |
2 |
1 |
3 |
1 |
3×5 |
1 |
2 |
1 |
3 |
1 |
5 |
1 |
5×7 |
1 |
2 |
1 |
5 |
1 |
7 |
∴
1 |
1×3 |
1 |
3×5 |
1 |
5×7 |
1 |
19×21 |
=
1 |
2 |
1 |
3 |
1 |
2 |
1 |
3 |
1 |
5 |
1 |
2 |
1 |
5 |
1 |
7 |
1 |
2 |
1 |
19 |
1 |
21 |
=
1 |
2 |
1 |
3 |
1 |
3 |
1 |
5 |
1 |
5 |
1 |
7 |
1 |
19 |
1 |
21 |
=
1 |
2 |
1 |
21 |
=
10 |
21 |
(1)
1 |
1×3 |
1 |
3×5 |
1 |
5×7 |
1 |
99×101 |
(2)利用类似方法,可求得:
1 |
1×4 |
1 |
4×7 |
1 |
7×10 |
1 |
19×22 |
(3)受以上启发,请你解下列方程:
1 |
x(x+3) |
1 |
(x+3)(x+6) |
1 |
(x+6)(x+9) |
3 |
x+9 |
分析:(1)根据上面的规律,将原式展开,再进行计算即可;
(2)由已知得
=
(1-
),依此类推,可得出
=
(
-
),再将原式展开计算即可;
(3)由上面的规律将原方程变形为
(
-
+
-
+
-
)=3×
.
(2)由已知得
1 |
1×4 |
1 |
3 |
1 |
4 |
1 |
19×22 |
1 |
3 |
1 |
19 |
1 |
22 |
(3)由上面的规律将原方程变形为
1 |
3 |
1 |
x |
1 |
x+3 |
1 |
x+3 |
1 |
x+6 |
1 |
x+6 |
1 |
x+9 |
1 |
x+9 |
解答:解:(1)原式=
(1-
)+
(
-
)+…+
(
-
)
=
(1-
+
-
+…+
-
)
=
(1-
)
=
×
=
;
(2)原式=
(1-
)+
(
-
)+…+
(
-
)
=
(1-
+
-
+…+
-
)
=
(1-
)
=
×
=
;
(3)原式可化为
(
-
+
-
+
-
)=3×
,
即
(
-
)=
,
解得x=1,
检验:把x=1代入x(x+9)=10≠0.
∴原方程的解为:x=1.
1 |
2 |
1 |
3 |
1 |
2 |
1 |
3 |
1 |
5 |
1 |
2 |
1 |
99 |
1 |
101 |
=
1 |
2 |
1 |
3 |
1 |
3 |
1 |
5 |
1 |
99 |
1 |
101 |
=
1 |
2 |
1 |
101 |
=
1 |
2 |
100 |
101 |
50 |
101 |
(2)原式=
1 |
3 |
1 |
4 |
1 |
3 |
1 |
4 |
1 |
7 |
1 |
3 |
1 |
19 |
1 |
22 |
=
1 |
3 |
1 |
4 |
1 |
4 |
1 |
7 |
1 |
19 |
1 |
22 |
=
1 |
3 |
1 |
22 |
=
1 |
3 |
21 |
22 |
=
7 |
22 |
(3)原式可化为
1 |
3 |
1 |
x |
1 |
x+3 |
1 |
x+3 |
1 |
x+6 |
1 |
x+6 |
1 |
x+9 |
1 |
x+9 |
即
1 |
3 |
1 |
x |
1 |
x+9 |
3 |
x+9 |
解得x=1,
检验:把x=1代入x(x+9)=10≠0.
∴原方程的解为:x=1.
点评:本题考查了解分式方程,先找出规律,在将方程整理是解此题的关键,注意分式方程要验根.
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