Câu 2:
a. 2cos2x + sinx = sin3x sin3x – sinx – 2cos2x = 0
2cos2xsinx – 2cos2x = 0 cos2x = 0 hay sinx = 1
(k Z)
x =
hay x = 2
2 k
4 k 2
b. log 2 (2x).log 3 (3x) > 1, đk x > 0
log 3 x + log 2 x + log 2 x.log 3 x > 0 log 3 2(log 2 x) 2 + (log 3 2 + 1)log 2 x > 0
log 2 x < -log 2 6 hay log 2 x > 0 0 < x < 1
6 hay x > 1
3
x dx
, đặt u = x 1 u 2 = x + 1 2udu = dx
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