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Giải các phương trình
a) \({{2({x^2} – 1)} \over {2x + 1}} = 2 – {{x + 2} \over {2x + 1}}\)
b) \({{2x – 5} \over {x – 1}} = {{5x – 3} \over {3x + 5}}\)
a) \({{2({x^2} – 1)} \over {2x + 1}} = 2 – {{x + 2} \over {2x + 1}}\)
Điều kiện: \(x \ne – {1 \over 2}\)
Ta có:
\(\eqalign{
& {{2({x^2} – 1)} \over {2x + 1}} = 2 – {{x + 2} \over {2x + 1}}\cr& \Leftrightarrow 2({x^2} – 1) = 2(2x + 1) – (x + 2) \cr
& \Leftrightarrow 2{x^2} – 2 = 4x + 2 – x – 2 \cr& \Leftrightarrow 2{x^2} – 3x – 2 = 0 \cr
& \Leftrightarrow \left[ \matrix{
x = 2 \;( \text{thỏa mãn})\hfill \cr
x = – {1 \over 2}\,(\text{loại} )\hfill \cr} \right. \cr} \)
Vậy S = {2}
b) \({{2x – 5} \over {x – 1}} = {{5x – 3} \over {3x + 5}}\)
Điều kiện:
\(\left\{ \matrix{
x \ne 1 \hfill \cr
x \ne – {5 \over 3} \hfill \cr} \right.\)
Ta có:
\(\eqalign{
& {{2x – 5} \over {x – 1}} = {{5x – 3} \over {3x + 5}}\cr& \Leftrightarrow (2x – 5)(3x + 5) = (5x – 3)(x – 1) \cr
& \Leftrightarrow 6{x^2} + 10x – 15 x- 25 = 5{x^2} – 5x – 3x + 3 \cr
& \Leftrightarrow {x^2} + 3x – 28 = 0 \Leftrightarrow \left[ \matrix{
x = 4\;( \text{thỏa mãn})\hfill \cr
x = – 7\;( \text{thỏa mãn}) \hfill \cr} \right. \cr} \)
Vậy S = {-7, 4}