Utf-8__logrit

  • April 2020
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Nguyễn Phú Khánh – Đà Lạt

http://www.toanthpt.net

Trong tài liệu mà tôi up dưới đây đa phần dạng toán của logarit như vầy log2x1(2x2 + x – 1) + logx+1(2x – 1)2 = 4. Điều kiện x >

1 và x  1 2

log 2x 1 (2x 2  x  1)  log x 1 (2x  1) 2  4

 log 2x 1 (x  1)(2x  1)  2 log x 1 (2x  1)  4  1  log 2x 1 (x  1)  2 log x 1 (2x  1)  4 

1  2log (x 1)  3 log x 1 (2x  1)

 2l og 2 x 1 (2x  1)  3log x 1 (2x  1)  1  0 .

Đây là pt bậc 2 theo log x 1 (2x  1) có a + b + c = 0 => log x 1 (2x  1)  1 và log x 1 (2x  1) 

1 . 2

* log x 1 (2x  1)  1  x + 1 = 2x − 1  x = 2 1  2x  1  x  1  (2x  1) 2  x  1  4x 2  5x  0 2 5  x = 0 (loại) hay x = . 4 5 Vậy : x = hay x = 2 4

* log x 1 (2x  1) 