1、解不等式(1/2)的x²+3>2的-4x 2、求方程log的2(x²-2x-3)=1+log的2
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1、解不等式(1/2)的x²+3>2的-4x 2、求方程log的2(x²-2x-3)=1+log的2(x+1)
你好!
(1/2)^(x²+3) > 2^(-4x) = (1/2)^(4x)
x² + 3 < 4x
x² - 4x + 3 < 0
(x - 1)(x - 3) < 0
1 < x < 3
log2 (x² - 2x - 3) = 1 + log2 (x+1)
log2 (x² - 2x - 3) = log2 (2) + log2 (x+1)
log2 (x² - 2x - 3) = log2 (2x+2)
x² - 2x - 3 = 2x + 2
x² - 4x - 5 = 0
(x+1)(x-5) = 0
∵x+1>0 (真数大于零)
∴x - 5 = 0
x = 5
(1/2)^(x²+3) > 2^(-4x) = (1/2)^(4x)
x² + 3 < 4x
x² - 4x + 3 < 0
(x - 1)(x - 3) < 0
1 < x < 3
log2 (x² - 2x - 3) = 1 + log2 (x+1)
log2 (x² - 2x - 3) = log2 (2) + log2 (x+1)
log2 (x² - 2x - 3) = log2 (2x+2)
x² - 2x - 3 = 2x + 2
x² - 4x - 5 = 0
(x+1)(x-5) = 0
∵x+1>0 (真数大于零)
∴x - 5 = 0
x = 5
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