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MAA AMC
AMC 12/AHSME
1978 AMC 12/AHSME
9
Simplify absolute value expression
Simplify absolute value expression
Source: 1978 AHSME Problem 9
May 31, 2014
absolute value
AMC
Problem Statement
If
x
<
0
x<0
x
<
0
, then
∣
x
−
(
x
−
1
)
2
∣
\left|x-\sqrt{(x-1)^2}\right|
x
−
(
x
−
1
)
2
equals
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
1
−
2
x
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
−
2
x
−
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
1
+
2
x
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
2
x
−
1
<span class='latex-bold'>(A) </span>1\qquad<span class='latex-bold'>(B) </span>1-2x\qquad<span class='latex-bold'>(C) </span>-2x-1\qquad<span class='latex-bold'>(D) </span>1+2x\qquad <span class='latex-bold'>(E) </span>2x-1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
1
−
2
x
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
−
2
x
−
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
1
+
2
x
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
2
x
−
1
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