问题
解答题
(1)一元二次方程x2-2x-
(2)先化简,再求值:
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答案
解方程x2-2x-
=0,得:x1=5 4
,x2=-5 2
,1 2
∵x2-(k+2)x+
=0,9 4
∴△=(k+2)2-9≥0,即k≥1或k≤-5,
①根据题意,把x=
代入x2-(k+2)x+5 2
=0,得:(9 4
)2-5 2
(k+2)+5 2
=0,9 4
解得:k=
;7 5
②把x=-
代入x2-(k+2)x+1 2
=0得:(-9 4
)2+1 2
(k+2)+1 2
=0,9 4
解得:k=-7,
综上所述,k的值为-7或
;7 5
(2)原式=
x•32 3
-2x2•x
+6x•x x2 x 2
=2x
-2x
+6xx x
=(8x-2)
,x
当x=4时,原式=(8×4-2)
=60.4