试题

题目:
计算题
(1)
32
-3
1
2
+
1
8
;               
(2)(
6
-
5
)(
6
+
5
)

(3)先化简再计算:
x2-1
x2+x
÷(x-
2x-1
x
)
,其中x是一元二次方程x2-2x-2=0的正数根.
答案
解:(1)原式=4
2
-3×
2
2
+
2
2
8

=4
2
-
3
2
2
+
2
4

=(4-
3
2
+
1
4
2

=
11
2
4

(2)原式=(
6
2-(
5
2
=6-5
=1;
(3)
x2-1
x2+x
÷(x-
2x-1
x
)

=
(x-1)(x+1)
x(x+1)
÷
x2-2x+1
x

=
(x-1)(x+1)
x(x+1)
·
x
(x-1)2

=
1
x-1


解x2-2x-2=0得,
x=
4-4×1×(-2)
2

x1=1+
3
,x2=1-
3

原式=
1
1+
3
-1
=
3
3

解:(1)原式=4
2
-3×
2
2
+
2
2
8

=4
2
-
3
2
2
+
2
4

=(4-
3
2
+
1
4
2

=
11
2
4

(2)原式=(
6
2-(
5
2
=6-5
=1;
(3)
x2-1
x2+x
÷(x-
2x-1
x
)

=
(x-1)(x+1)
x(x+1)
÷
x2-2x+1
x

=
(x-1)(x+1)
x(x+1)
·
x
(x-1)2

=
1
x-1


解x2-2x-2=0得,
x=
4-4×1×(-2)
2

x1=1+
3
,x2=1-
3

原式=
1
1+
3
-1
=
3
3
考点梳理
分式的化简求值;二次根式的混合运算;解一元二次方程-公式法.
(1)先化为最简二次根式再加减;
(2)根据平方差公式进行计算;
(3)先化简分式,再计算一元二次方程的根,代入计算即可.
本题考查了分式的化简求值、二次根式的混合运算,会因式分解是解题的关键.
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