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Giải các phương trình
a) \(\sqrt {2x + 8} = 3x + 4\)
b) |x2 + 5x + 6| = 3x + 13
c) (x2 + 3x)(x2 + 3x + 4) = 5
Đáp án
a) Ta có:
\(\eqalign{
& \sqrt {2x + 8} = 3x + 4 \Leftrightarrow \left\{ \matrix{
3x + 4 \ge 0 \hfill \cr
2x + 8 = {(3x + 4)^2} \hfill \cr} \right. \cr&\Leftrightarrow \left\{ \matrix{
x \ge – {4 \over 3} \hfill \cr
9{x^2} + 22x – 8 = 0 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
x \ge – {4 \over 3} \hfill \cr
\left[ \matrix{
x = 2\;(\text{ loại}) \hfill \cr
x = – {4 \over 3} \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow x = – {4 \over 9} \cr} \)
Vậy \(S = {\rm{\{ }} – {4 \over 9}{\rm{\} }}\)
b) Điều kiện: \(3x + 13 \ge 0 \Leftrightarrow x \ge – {{13} \over 3}\)
Ta có:
\(\eqalign{
& |{x^2} + 5x + 6| = 3x + 13 \cr&\Leftrightarrow \left[ \matrix{
{x^2} + 5x + 6 = 3x + 13 \hfill \cr
{x^2} + 5x + 6 = – (3x + 13) \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
{x^2} + 2x – 7 = 0 \hfill \cr
{x^2} + 8x + 19 = 0 \hfill \cr} \right. \Leftrightarrow x = – 1 \pm 2\sqrt 2 \cr} \)
Vậy \(S = {\rm{\{ }} – 1 – 2\sqrt 2 ;\, – 1 + 2\sqrt 2 {\rm{\} }}\)
c) Đặt t = x2+ 3x, ta có phương trình:
\(\eqalign{
& t(t + 4) = 5 \Leftrightarrow {t^2} + 4t – 5 = 0 \Leftrightarrow \left[ \matrix{
t = 1 \hfill \cr
t = – 5 \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
{x^2} + 3x – 1 = 0 \hfill \cr
{x^2} + 3x + 5 = 0 \hfill \cr} \right. \Leftrightarrow x = {{ – 3 \pm \sqrt {13} } \over 2} \cr} \)
Vậy \(S = {\rm{\{ }}{{ – 3 \pm \sqrt {13} } \over 2}{\rm{\} }}\)