log2(8-2x-x^2)≤3 ~~~~~~~~~~~~~~~~~~~~~~要求详细过程

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log2(8-2x-x^2)≤3 ~~~~~~~~~~~~~~~~~~~~~~要求详细过程
log2(8-2x-x^2)≤3 ~~~~~~~~~~~~~~~~~~~~~~
要求详细过程

log2(8-2x-x^2)≤3 ~~~~~~~~~~~~~~~~~~~~~~要求详细过程
3=log2(8)
底数2大于1,log2(x)是增函数
所以8-2x-x^2=0
x(x+2)>=0
x=0
真数大于0
8-2x-x^2〉0
x^2+2x-8

8-2x-x^2≤2^3=8
x^2+2x>=0
x(x+2)>=0
x>=0或X<=-2

将式子化为8-2x-x^2≤2^3，则可化为-2x-x^2≤0，求得x>=0或者x<=-2。
并且8-2x-x^2>0故求得:-4综上所述可知0<=x<2或-4

8-2x-x^2<=8,8-2x-x^2>0
x^2+2x>=0
x>=0或x<=-2
8-2x-x^2>0
x^2+2x-8<0
-4所以-4

log2(8-2x-x^2)≤3=log2(2^3)
所以有:
0<8-2x-x^2<=8
(i)8-2x-x^2>0
x^2+2x-8<0
(x+4)(x-2)<0
解得:-4(ii)8-2x-x^2<=8
x^2+2x>=0
x(x+2)>=0
解得:x>=0或x<=-2
综上所述,0<=x<2或-4

8-2x-x^2<=9
x^2+2x+1>=0
(x+1)^2>=0
所以x任意值
但是
8-2x-x^2>0
-4

0<8-2X-X^2<=8
解这个不等式,可得x属于[-4,-2)并(0,2]

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