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> (1)计算tan30°•cot60°+sin30°-22sin45°+(1-cot30°)0(2)解方程:(2x+1)2-3(2x+1)=4(3)化简:11+2+12+3+…+18+9.-数学
(1)计算tan30°•cot60°+sin30°-22sin45°+(1-cot30°)0(2)解方程:(2x+1)2-3(2x+1)=4(3)化简:11+2+12+3+…+18+9.-数学
题目简介
(1)计算tan30°•cot60°+sin30°-22sin45°+(1-cot30°)0(2)解方程:(2x+1)2-3(2x+1)=4(3)化简:11+2+12+3+…+18+9.-数学
题目详情
(1)计算tan30°•cot60°+sin30°-
2
2
sin45°+(1-cot30°)
0
(2)解方程:(2x+1)
2
-3(2x+1)=4
(3)化简:
1
1+
2
+
1
2
+
3
+…+
1
8
+
9
.
题型:解答题
难度:中档
来源:不详
答案
(1)原式=
3
3
×
3
3
+
class="stub"1
2
-
2
2
×
2
2
+1,
=
class="stub"1
3
+
class="stub"1
2
-
class="stub"1
2
+1,
=
class="stub"4
3
;
(2)设2x+1=t,则由原方程,得
t2-3t=4,即(t-4)(t+1)=0,
解得,t=4或t=-1;
①当t=4时,2x+1=4,
解得x=
class="stub"3
2
;
②当t=-1时,2x+1=-1,
解得,x=-1;
综上所述,原方程的解为x=
class="stub"3
2
,或x=-1;
(3)原式=
1-
2
1-2
+
2
-
3
2-3
+…+
8
-
9
8-9
=-(1-
2
+
2
-
3
+…+
8
-
9
)=-(1-3)=2.
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题目简介
(1)计算tan30°•cot60°+sin30°-22sin45°+(1-cot30°)0(2)解方程:(2x+1)2-3(2x+1)=4(3)化简:11+2+12+3+…+18+9.-数学
题目详情
(2)解方程:(2x+1)2-3(2x+1)=4
(3)化简:
答案
=
=
(2)设2x+1=t,则由原方程,得
t2-3t=4,即(t-4)(t+1)=0,
解得,t=4或t=-1;
①当t=4时,2x+1=4,
解得x=
②当t=-1时,2x+1=-1,
解得,x=-1;
综上所述,原方程的解为x=
(3)原式=