Question

求解答这两道高中数学题

Answer 1

2x^2_4x_16<0,
x^2_2x_8<0,
(x_4)(x+2)<0,
_2<x<4

追答

x^2_2x_8≥(m+2)x_m_15
x^2_(4+m)x+m+7≥0,
x>2是它的解，
不等式可表示为(x_2)[x_(m+7)/2]≥0,
x>2,x>(m+7)/2且2>(m+7)/2,
4>m+7,m<3

追问

19题呢

追答

当a>0时，g(x)=ax^2-2ax+1+b为增函数，有g(3)=9a-6a+1+b=4; g(2)=4a-4a+1+b=1; 故b=0, a=1
当a>0时，g(x)=ax^2-2ax+1+b为减函数，有g(3)=9a-6a+1+b=1; g(2)=4a-4a+1+b=3; 故b=3, a=-1（舍去）
b=0, a=1
此时g(x)=x^2-2x+1, f(x)=x-2+1/x,
f(2^x)=2^x-2+1/2^x,方程f(2^x)-k*2^x=2^x-2+1/2^x-k*2^x>=0,
当f(2^x)-k*2^x=2^x-2+1/2^x-k*2^x=0, (2^x)^2-2*2^x+1-k*(2^x)^2=0,
(2^x-1)^2=k(2^x)^2,
因-1<=x<=1,有1/2<=2^x<=2, -1/2<=2^x-1<=1, 0<=(2^x-1)^2<=1,
0<=k(2^x)^2<=1
因1/2<=2^x<=2，1/4<=(2^x)^2<=4，0<=k<=1/4