(2x+1)ⁿ=0成立的实数x的范围是(   )

A.x=-              B.-＜x＜0

C.-1＜x＜0           D.-1＜x≤0

已知a，b，c∈R，函数f(x)＝ax2＋bx＋c.若f(0)＝f(4)＞f(1)，则 (　　)．

A．a＞0,4a＋b＝0 B．a＜0,4a＋b＝0

C．a＞0,2a＋b＝0 D．a＜0,2a＋b＝0

A.p-q=0                           B.p+q=0

C.p+q=-5                        D.2p+q=-4

A.x²+y²-2x-y+=0                          B.x²+y²+2x-y+1=0

C.x²+y²+2x-y+=0                          D.x²+y²-2x+y+1=0