Hence, or otherwise, solve [2^x/(2^x - 2)] <= 2
thanks
更新1:
is this question out c?? thz
2001 ce amaths (2)
2009-04-04 5:53 pm
Solve [y/(y - 2)] <= 2
Hence, or otherwise, solve [2^x/(2^x - 2)] <= 2
thanks
Hence, or otherwise, solve [2^x/(2^x - 2)] <= 2
thanks
回答 (2)
2009-04-04 7:22 pm
✔ 最佳答案
y/(y-2) < = 2y(y-2)^2/(y-2) < = 2(y-2)^2
y(y-2) < = 2(y-2)^2
y^2 - 2y < = 2y^2 + 8 - 8y
0 < = y^2 - 6y + 8
0 < = (y-2)(y- 4)
so y < 2 or y> = 4. (y cannot be equal to 2 since y-2 = 0).
2^x <= 2^1
so x <= 1
and 2^x => 4
2^x => 2^2
x => 2
2009-04-04 10:25 pm
y < 2 or y >= 4
2^x < 2 or 2^x => 4
2^x < 2^1 or 2^x => 2^2
x < 1 or x => 2
(x<>1)
2^x < 2 or 2^x => 4
2^x < 2^1 or 2^x => 2^2
x < 1 or x => 2
(x<>1)
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