已知实数a、b互为相反数,c、d互为倒数,x的绝对值为7,求代数式x2+(a+b+cd)x+a+b+3cd的值

2025-05-12 13:34:19
推荐回答(1个)
回答1:

根据题意,得
a+b=0  ①
cd=1   ②
|x|=

7
,即x=±
7

(1)当x=
7
时,
x2+(a+b+cd)x+
a+b
+
3 cd

=(
7
)2
+(0+1)×
7
+0+1,
=7+
7
+1,
=8+
7

(2)当x=-
7
时,
x2+(a+b+cd)x+
a+b
+
3 cd

=(?
7
)2
+(0+1)×(?
7
)+0+1,
=7-
7
+1,
=8-
7

所以,代数式x2+(a+b+cd)x+
a+b
+
3 cd
的值是8±
7