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(1)计算:2−1+3cos30°+|−5|−(π−2011)0.(2)解分式方程:2xx+1+3x−1=2.
题目详情
(1)计算:2−1+
cos30°+|−5|−(π−2011)0.
(2)解分式方程:
+
=2.2−1+
cos30°+|−5|−(π−2011)0.
(2)解分式方程:
+
=2.2−1+
cos30°+|−5|−(π−2011)0.
(2)解分式方程:
+
=2.2−1+
cos30°+|−5|−(π−2011)0.
(2)解分式方程:
+
=2.−1+
cos30°+|−5|−(π−2011)0.
(2)解分式方程:
+
=2.
3 3 )0.
(2)解分式方程:
+
=2.)0.
(2)解分式方程:
+
=2.0.
(2)解分式方程:
+
=2.
+
=2.
2x 2x x+1 x+1
3 3 x−1 x−1
3 |
(2)解分式方程:
2x |
x+1 |
3 |
x−1 |
3 |
(2)解分式方程:
2x |
x+1 |
3 |
x−1 |
3 |
(2)解分式方程:
2x |
x+1 |
3 |
x−1 |
3 |
(2)解分式方程:
2x |
x+1 |
3 |
x−1 |
3 |
(2)解分式方程:
2x |
x+1 |
3 |
x−1 |
3 |
(2)解分式方程:
2x |
x+1 |
3 |
x−1 |
(2)解分式方程:
2x |
x+1 |
3 |
x−1 |
(2)解分式方程:
2x |
x+1 |
3 |
x−1 |
2x |
x+1 |
3 |
x−1 |
2x |
x+1 |
3 |
x−1 |
▼优质解答
答案和解析
(1)原式=
+
×
+5-1=6;
(2)去分母得:2x(x-1)+3(x+1)=2(x+1)(x-1),
去括号得:2x2-2x+3x+3=2x2-2,
移项合并得:x=-5,
经检验x=-5是分式方程的解.
1 1 12 2 2+
×
+5-1=6;
(2)去分母得:2x(x-1)+3(x+1)=2(x+1)(x-1),
去括号得:2x2-2x+3x+3=2x2-2,
移项合并得:x=-5,
经检验x=-5是分式方程的解.
3 3 3×
+5-1=6;
(2)去分母得:2x(x-1)+3(x+1)=2(x+1)(x-1),
去括号得:2x2-2x+3x+3=2x2-2,
移项合并得:x=-5,
经检验x=-5是分式方程的解.
3 3 32 2 2+5-1=6;
(2)去分母得:2x(x-1)+3(x+1)=2(x+1)(x-1),
去括号得:2x22-2x+3x+3=2x22-2,
移项合并得:x=-5,
经检验x=-5是分式方程的解.
1 |
2 |
3 |
| ||
2 |
(2)去分母得:2x(x-1)+3(x+1)=2(x+1)(x-1),
去括号得:2x2-2x+3x+3=2x2-2,
移项合并得:x=-5,
经检验x=-5是分式方程的解.
1 |
2 |
3 |
| ||
2 |
(2)去分母得:2x(x-1)+3(x+1)=2(x+1)(x-1),
去括号得:2x2-2x+3x+3=2x2-2,
移项合并得:x=-5,
经检验x=-5是分式方程的解.
3 |
| ||
2 |
(2)去分母得:2x(x-1)+3(x+1)=2(x+1)(x-1),
去括号得:2x2-2x+3x+3=2x2-2,
移项合并得:x=-5,
经检验x=-5是分式方程的解.
| ||
2 |
3 |
3 |
3 |
(2)去分母得:2x(x-1)+3(x+1)=2(x+1)(x-1),
去括号得:2x22-2x+3x+3=2x22-2,
移项合并得:x=-5,
经检验x=-5是分式方程的解.
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