Advertisements (Quảng cáo)
Giải các phương trình sau:
a) \(x + {1 \over {x – 1}} = {{2x – 1} \over {x – 1}}\)
b) \(x + {1 \over {x – 2}} = {{2x – 3} \over {x – 2}}\)
c) \(({x^2} – 3x + 2)\sqrt {x – 3} = 0\)
d) \(({x^2} – x – 2)\sqrt {x + 1} = 0\)
a) ĐKXĐ: \(x ≠ 1\)
Ta có:
\(\eqalign{
& x + {1 \over {x – 1}} = {{2x – 1} \over {x – 1}} \Leftrightarrow x(x – 1) + 1 = 2x – 1 \cr
& \Leftrightarrow {x^2} – 3x + 2 = 0 \Leftrightarrow \left[ \matrix{
x = 1\,(\text{loại}) \hfill \cr
x = 2 \hfill \cr} \right. \cr} \)
Vậy S = {2}
b) ĐKXĐ: \(x ≠ 2\)
Ta có:
\(\eqalign{
& x + {1 \over {x – 2}} = {{2x – 3} \over {x – 2}} \Leftrightarrow {x^2} – 2x + 1 = 2x – 3 \cr
& \Leftrightarrow {x^2} – 4x + 4 = 0 \Leftrightarrow {(x – 2)^2} = 0 \cr
& \Leftrightarrow x = 2\,(\text{loại}) \cr} \)
Advertisements (Quảng cáo)
Vậy S = Ø
c) ĐKXĐ: \(x ≥ 3\)
Ta có:
\(\eqalign{
& ({x^2} – 3x + 2)\sqrt {x – 3} = 0 \Leftrightarrow \left[ \matrix{
\sqrt {x – 3} = 0 \hfill \cr
{x^2} – 3x + 2 = 0 \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = 3 \hfill \cr
x = 1\,(\text{loại}) \hfill \cr
x = 2\,(\text{loại}) \hfill \cr} \right. \cr} \)
Vậy S = {3}
d) ĐKXĐ: \(x ≥ -1\)
Ta có:
\(({x^2} – x – 2)\sqrt {x + 1} = 0 \Leftrightarrow \left[ \matrix{
\sqrt {x + 1} = 0 \hfill \cr
{x^2} – x – 2 = 0 \hfill \cr} \right.\)
\(\Leftrightarrow \left[ \matrix{
x = – 1 \hfill \cr
x = 2 \hfill \cr} \right.\)
Vậy S = {-1, 2}