题目内容
规定新运算符号*的运算过程为a*b=
a-
b,则
(1)求5*(-5);
(2)解方程2*(2*x)=1*x.
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(1)求5*(-5);
(2)解方程2*(2*x)=1*x.
分析:(1)根据新定义运算得到5*(-5)=
×5-
×(-5),然后进行实数的加减运算;
(2)先根据新定义得到2*(
-
)=
-
(
-
)=
-
+
=
+
,1*x=
-
,则
+
=
-
,再去分母得到24+3x=16-12x,移项得到15x=-8,然后把x的系数化为1即可.
| 1 |
| 3 |
| 1 |
| 4 |
(2)先根据新定义得到2*(
| 2 |
| 3 |
| x |
| 4 |
| 2 |
| 3 |
| 1 |
| 4 |
| 2 |
| 3 |
| x |
| 4 |
| 2 |
| 3 |
| 1 |
| 6 |
| x |
| 16 |
| 1 |
| 2 |
| x |
| 16 |
| 1 |
| 3 |
| x |
| 4 |
| 1 |
| 2 |
| x |
| 16 |
| 1 |
| 3 |
| x |
| 4 |
解答:解:(1)5*(-5)=
×5-
×(-5)=
+
=
;
(2)∵2*x=
-
,
∴2*(
-
)=
-
(
-
)=
-
+
=
+
1*x=
-
,
∴
+
=
-
,
去分母得,24+3x=16-12x,
移项得,15x=-8,
系数化为1得,x=-
.
| 1 |
| 3 |
| 1 |
| 4 |
| 5 |
| 3 |
| 5 |
| 4 |
| 35 |
| 12 |
(2)∵2*x=
| 2 |
| 3 |
| x |
| 4 |
∴2*(
| 2 |
| 3 |
| x |
| 4 |
| 2 |
| 3 |
| 1 |
| 4 |
| 2 |
| 3 |
| x |
| 4 |
| 2 |
| 3 |
| 1 |
| 6 |
| x |
| 16 |
| 1 |
| 2 |
| x |
| 16 |
1*x=
| 1 |
| 3 |
| x |
| 4 |
∴
| 1 |
| 2 |
| x |
| 16 |
| 1 |
| 3 |
| x |
| 4 |
去分母得,24+3x=16-12x,
移项得,15x=-8,
系数化为1得,x=-
| 8 |
| 15 |
点评:本题考查了解一元一次方程:先去分母,再去括号,然后把含未知数的项移到方程左边,常数移到方程右边,再合并,把未知数的系数化为1即可.也考查了对新定义的理解能力.
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