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Giải các phương trình sau
a) \(|x^2– 2x – 3| = 2x + 2\)
b) \(\sqrt {{x^2} – 4} = 2(x – \sqrt 3 )\)
Đáp án
a) Điều kiện: \(x ≥ -1\). Ta có:
\(\eqalign{
& \left| {{x^2}-2x-3} \right| = 2x + {\rm{ }}2\cr& \Leftrightarrow \left[ \matrix{
{x^2}-2x-3 = 2x + 2 \hfill \cr
{x^2}-2x-3 = – 2x – 2 \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
{x^2} – 4x – 5 = 0 \hfill \cr
{x^2} – 1 = 0 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = – 1;\,x = 5 \hfill \cr
x = \pm 1 \hfill \cr} \right. (\text{nhận})\cr} \)
Vậy S = {-1, 1, 5}
b) Ta có:
\(\sqrt {{x^2} – 4} = 2(x – \sqrt 3 )\)
\(\Leftrightarrow \left\{ \matrix{
x \ge \sqrt 3 \hfill \cr
{x^2} – 4 = 4({x^2} – 2\sqrt 3 + 3) \hfill \cr} \right. \)
\(\Leftrightarrow \left\{ \matrix{
x \ge \sqrt 3 \hfill \cr
3{x^2} – 8\sqrt 3 + 16 = 0 \hfill \cr} \right.\)
Vậy \(S = {\rm{\{ }}{{4\sqrt 3 } \over 3}{\rm{\} }}\)