4^X-3×2^X+2+32
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4^X-3×2^X+2+324^X-3×2^X+2+324^X-3×2^X+2+32令a=2^x则4^x=a²2^(x+2)=2^x*2^2=4a所以a²-3*4a+324^X-3
4^X-3×2^X+2+32
4^X-3×2^X+2+32
4^X-3×2^X+2+32
令a=2^x
则4^x=a²
2^(x+2)=2^x*2^2=4a
所以a²-3*4a+32
4^X-3×2^X+2+32<0
(2^X)^2-12×2^X+32<0
(2^x-4)(2^x-8)<0
4<2^x<8
2
4^X-3*2^(X+2)+32<0
(2^X)平方-12*2^X+32<0
(2^X-4)(2^X-8)<0
4<2^X<8
2
你的式子应该是:
4^X-3×2^(X+2)+32<0
设2^x为y,则4^x为y^2,原式=y^2-12y+32<0
解出y的取值范围后,再用y=2^x去解x的范围
*-----------------------------------------------*| 6 4 X | 8 X X | X X 5 || X X X | X X X | X 7 8 || X X X | X X X | X X X ||---------------+---------------+--------------- || X X X | X X X | 5 1 X || X X X | X 6 X | X X X || 8 X X | 3 5 X | 2 X X ||
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