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> (1)计算:2-1+(2π-1)0-22sin45°-3tan30°(2)解方程:(x-8)(x-1)=-12(3)一个不透明的布袋里装有4个大小、质地均相同的乒乓球,每个球上面分别标有1、2、3、4
(1)计算:2-1+(2π-1)0-22sin45°-3tan30°(2)解方程:(x-8)(x-1)=-12(3)一个不透明的布袋里装有4个大小、质地均相同的乒乓球,每个球上面分别标有1、2、3、4
题目简介
(1)计算:2-1+(2π-1)0-22sin45°-3tan30°(2)解方程:(x-8)(x-1)=-12(3)一个不透明的布袋里装有4个大小、质地均相同的乒乓球,每个球上面分别标有1、2、3、4
题目详情
(1)计算:2
-1
+(2π-1)
0
-
2
2
sin45°-
3
tan30°
(2)解方程:(x-8)(x-1)=-12
(3)一个不透明的布袋里装有4个大小、质地均相同的乒乓球,每个球上面分别标有1、2、3、4,小林先从布袋中随机抽取一个乒乓球(不放回去),再从剩下的3个球中随机抽取第二个乒乓球.
①请你列出所有可能的结果;
②求两次取得乒乓球的数字之积为奇数的概率.
题型:解答题
难度:中档
来源:不详
答案
(1)原式=
class="stub"1
2
+1-
2
2
×
2
2
-
3
×
3
3
=
class="stub"1
2
+1-
class="stub"1
2
-1
=0;
(2)原方程变形得:x2-9x+20=0
方程左边分解因式得:(x-4)(x-5)=0
∴x-4=0或x-5=0,
∴x1=4,x2=5;
(3)①根据题意列表如下:
1
2
3
4
1
(1,2)
(1,3)
(1,4)
2
(2,1)
(2,3)
(2,4)
3
(3,1)
(3,2)
(3,4)
4
(4,1)
(4,2)
(4,3)
由以上表格可知:有12种可能结果
②在①中的12种可能结果中,两个数字之积为奇数的只有2种,
∴P(两个数字之积是奇数)=
class="stub"2
12
=
class="stub"1
6
.
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题目简介
(1)计算:2-1+(2π-1)0-22sin45°-3tan30°(2)解方程:(x-8)(x-1)=-12(3)一个不透明的布袋里装有4个大小、质地均相同的乒乓球,每个球上面分别标有1、2、3、4
题目详情
(2)解方程:(x-8)(x-1)=-12
(3)一个不透明的布袋里装有4个大小、质地均相同的乒乓球,每个球上面分别标有1、2、3、4,小林先从布袋中随机抽取一个乒乓球(不放回去),再从剩下的3个球中随机抽取第二个乒乓球.
①请你列出所有可能的结果;
②求两次取得乒乓球的数字之积为奇数的概率.
答案
=
=0;
(2)原方程变形得:x2-9x+20=0
方程左边分解因式得:(x-4)(x-5)=0
∴x-4=0或x-5=0,
∴x1=4,x2=5;
(3)①根据题意列表如下:
②在①中的12种可能结果中,两个数字之积为奇数的只有2种,
∴P(两个数字之积是奇数)=