1. 下列方程中,不是分式方程的是 (
A.$\frac{1}{x}=2$;
B.$\frac{1}{x+2}=3$;
C.$\frac{x+1}{3}+\frac{x-1}{2}=1$;
D.$\frac{1}{x+2}=\frac{3}{x}$。
C
)A.$\frac{1}{x}=2$;
B.$\frac{1}{x+2}=3$;
C.$\frac{x+1}{3}+\frac{x-1}{2}=1$;
D.$\frac{1}{x+2}=\frac{3}{x}$。
答案
C.
2. 解分式方程$\frac{3x}{x-1} - 2 = \frac{2}{1-x}$,去分母得 (
A.$3x - 2(x - 1) = -2$;
B.$3x - 2(x - 1) = 2$;
C.$3x - 2x - 2 = -2$;
D.$3x - 2x + 2 = 2$;
A
)A.$3x - 2(x - 1) = -2$;
B.$3x - 2(x - 1) = 2$;
C.$3x - 2x - 2 = -2$;
D.$3x - 2x + 2 = 2$;
答案
A.
3. 解方程:
(1) $\frac{3}{x+3}=4$;
(2) $\frac{7}{3x-1}=\frac{2}{x}$;
(3) $\frac{2-x}{3x-1}+\frac{7}{1-3x}=1$;
(4) $\frac{5x-4}{2x-4}=\frac{2x+5}{3x-6}-\frac{1}{2}$。
(1) $\frac{3}{x+3}=4$;
(2) $\frac{7}{3x-1}=\frac{2}{x}$;
(3) $\frac{2-x}{3x-1}+\frac{7}{1-3x}=1$;
(4) $\frac{5x-4}{2x-4}=\frac{2x+5}{3x-6}-\frac{1}{2}$。
答案
(1)解:方程两边同乘(x + 3),得3 = 4(x + 3),
3 = 4x + 12,
4x = 3 - 12,
4x = -9,
$x = -\dfrac{9}{4}$,
检验:当$x = -\dfrac{9}{4}$时,$x + 3 = -\dfrac{9}{4} + 3 = \dfrac{3}{4} \neq 0$,
所以$x = -\dfrac{9}{4}$是原方程的解.
(2)解:方程两边同乘x(3x - 1),得7x = 2(3x - 1),
7x = 6x - 2,
7x - 6x = -2,
x = -2,
检验:当x = -2时,$x(3x - 1) = -2×(-6 - 1) = -2×(-7) = 14 \neq 0$,
所以x = -2是原方程的解.
(3)解:原方程可化为$\dfrac{2 - x}{3x - 1} - \dfrac{7}{3x - 1} = 1$,
方程两边同乘(3x - 1),得2 - x - 7 = 3x - 1,
-x - 5 = 3x - 1,
-x - 3x = -1 + 5,
-4x = 4,
x = -1,
检验:当x = -1时,$3x - 1 = -3 - 1 = -4 \neq 0$,
所以x = -1是原方程的解.
(4)解:方程两边同乘6(x - 2)(因为2x - 4 = 2(x - 2),3x - 6 = 3(x - 2)),得3(5x - 4) = 2(2x + 5) - 3(x - 2),
15x - 12 = 4x + 10 - 3x + 6,
15x - 12 = x + 16,
15x - x = 16 + 12,
14x = 28,
x = 2,
检验:当x = 2时,6(x - 2) = 0,
所以x = 2是增根,原方程无解.
3 = 4x + 12,
4x = 3 - 12,
4x = -9,
$x = -\dfrac{9}{4}$,
检验:当$x = -\dfrac{9}{4}$时,$x + 3 = -\dfrac{9}{4} + 3 = \dfrac{3}{4} \neq 0$,
所以$x = -\dfrac{9}{4}$是原方程的解.
(2)解:方程两边同乘x(3x - 1),得7x = 2(3x - 1),
7x = 6x - 2,
7x - 6x = -2,
x = -2,
检验:当x = -2时,$x(3x - 1) = -2×(-6 - 1) = -2×(-7) = 14 \neq 0$,
所以x = -2是原方程的解.
(3)解:原方程可化为$\dfrac{2 - x}{3x - 1} - \dfrac{7}{3x - 1} = 1$,
方程两边同乘(3x - 1),得2 - x - 7 = 3x - 1,
-x - 5 = 3x - 1,
-x - 3x = -1 + 5,
-4x = 4,
x = -1,
检验:当x = -1时,$3x - 1 = -3 - 1 = -4 \neq 0$,
所以x = -1是原方程的解.
(4)解:方程两边同乘6(x - 2)(因为2x - 4 = 2(x - 2),3x - 6 = 3(x - 2)),得3(5x - 4) = 2(2x + 5) - 3(x - 2),
15x - 12 = 4x + 10 - 3x + 6,
15x - 12 = x + 16,
15x - x = 16 + 12,
14x = 28,
x = 2,
检验:当x = 2时,6(x - 2) = 0,
所以x = 2是增根,原方程无解.
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