Bài 84 trang 156 SGK Đại số 10 nâng cao

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 \(S = {\rm{\{ }}{{4\sqrt 3 } \over 3}{\rm{\} }}\)