Bài 39. Giải phương trình bằng cách đưa về phương trình tích.
a) \((3{x^{2}} - {\rm{ }}7x{\rm{ }}-{\rm{ }}10)[2{x^2} + {\rm{ }}\left( {1{\rm{ }} - {\rm{ }}\sqrt 5 } \right)x{\rm{ }} + {\rm{ }}\sqrt 5 {\rm{ }}-{\rm{ }}3]{\rm{ }} = {\rm{ }}0\);
b) \({x^3} + {\rm{ }}3{x^2}-{\rm{ }}2x{\rm{ }}-{\rm{ }}6{\rm{ }} = {\rm{ }}0\);
c) \(({x^{2}} - {\rm{ }}1)\left( {0,6x{\rm{ }} + {\rm{ }}1} \right){\rm{ }} = {\rm{ }}0,6{x^2} + {\rm{ }}x\);
d) \({({x^2} + {\rm{ }}2x{\rm{ }}-{\rm{ }}5)^2} = {\rm{ }}{({\rm{ }}{x^2}-{\rm{ }}x{\rm{ }} + {\rm{ }}5)^2}\).
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a) \((3{x^{2}} - {\rm{ }}7x{\rm{ }}-{\rm{ }}10)[2{x^2} + {\rm{ }}\left( {1{\rm{ }} - {\rm{ }}\sqrt 5 } \right)x{\rm{ }} + {\rm{ }}\sqrt 5 {\rm{ }}-{\rm{ }}3]{\rm{ }} = {\rm{ }}0\)
\(\Leftrightarrow\)\(\left[ \matrix{
(3{x^{2}} - {\rm{ }}7x{\rm{ }}-{\rm{ }}10){\rm{ }} = {\rm{ }}0(1) \hfill \cr
2{x^2} + {\rm{ }}\left( {1{\rm{ }} - {\rm{ }}\sqrt 5 } \right)x{\rm{ }} + \sqrt 5 -{\rm{ }}3{\rm{ }} = {\rm{ }}0(2) \hfill \cr} \right.\)
Giải (1): phương trình \(a - b + c = 3 + 7 - 10 = 0\)
nên \({x_1} = - 1,{x_2} = - {{ - 10} \over 3} = {{10} \over 3}\)
Giải (2): phương trình có \(a + b + c = 2 + (1 - \sqrt{5}) + \sqrt{5} - 3 = 0\)
nên \({x_3} = 1,{x_4} = {{\sqrt 5 - 3} \over 2}\)
b) \({x^3} + {\rm{ }}3{x^2}-{\rm{ }}2x{\rm{ }}-{\rm{ }}6{\rm{ }} = {\rm{ }}0\) \(\Leftrightarrow {x^2}\left( {x{\rm{ }} + {\rm{ }}3} \right){\rm{ }}-{\rm{ }}2\left( {x{\rm{ }} + {\rm{ }}3} \right){\rm{ }} = {\rm{ }}0 \)
\(\Leftrightarrow \left( {x{\rm{ }} + {\rm{ }}3} \right)({x^2} - {\rm{ }}2){\rm{ }} = {\rm{ }}0\)
\(\Leftrightarrow\)\(\left[ \matrix{
x + 3 = 0 \hfill \cr
{x^2} - {\rm{ }}2{\rm{ }} = {\rm{ }}0 \hfill \cr} \right.\)
Giải ra \({x_1} = {\rm{ }} - 3,{\rm{ }}{x_{2}} = {\rm{ }} - \sqrt 2 ,{\rm{ }}{x_{3}} = \sqrt 2 \)
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c) \(({x^{2}} - {\rm{ }}1)\left( {0,6x{\rm{ }} + {\rm{ }}1} \right){\rm{ }} = {\rm{ }}0,6{x^2} + {\rm{ }}x\) \( \Leftrightarrow {\rm{ }}\left( {0,6x{\rm{ }} + {\rm{ }}1} \right)\left( {{x^2}-{\rm{ }}x{\rm{ }}-{\rm{ }}1} \right){\rm{ }} = {\rm{ }}0\)
\(\Leftrightarrow \left[ \matrix{
0,6x + 1 = 0(1) \hfill \cr
{x^2}-{\rm{ }}x{\rm{ }}-{\rm{ }}1{\rm{ }} = {\rm{ }}0(2) \hfill \cr} \right.\)
(1) ⇔ \(0,6x + 1 = 0 \)
\( \Leftrightarrow {x_1} = - {1 \over {0,6}} = - {5 \over 3}\)
(2):\(\Delta = {( - 1)^2} - 4.1.( - 1) = 1 + 4 = 5,\sqrt \Delta = \sqrt 5,\)
\({x_2} = {\rm{ }}{{1 - \sqrt 5 } \over 2},{x_3} = {{1 + \sqrt 5 } \over 2}\)
Vậy phương trình có ba nghiệm:
\({x_1} = - {5 \over 3},{x_2} = {{1 - \sqrt 5 } \over 2},{x_3} = {{1 + \sqrt 5 } \over 2}\),
d) \({({x^2} + {\rm{ }}2x{\rm{ }}-{\rm{ }}5)^2} = {\rm{ }}{({\rm{ }}{x^2}-{\rm{ }}x{\rm{ }} + {\rm{ }}5)^2}\)\( \Leftrightarrow {\rm{ }}{({x^2} + {\rm{ }}2x{\rm{ }}-{\rm{ }}5)^2} - {\rm{ }}{({\rm{ }}{x^2}-{\rm{ }}x{\rm{ }} + {\rm{ }}5)^2} = {\rm{ }}0\)
\(\Leftrightarrow ({x^2} + {\rm{ }}2x{\rm{ }}-{\rm{ }}5{\rm{ }} + {\rm{ }}{x^2}-{\rm{ }}x{\rm{ }} + {\rm{ }}5).\)
\(({\rm{ }}{x^2} + {\rm{ }}2x{\rm{ }}-{\rm{ }}5{\rm{ }} - {\rm{ }}{x^2} + {\rm{ }}x{\rm{ }} - {\rm{ }}5){\rm{ }} = {\rm{ }}0\)
\( \Leftrightarrow {\rm{ }}(2{x^2} + {\rm{ }}x)\left( {3x{\rm{ }}-{\rm{ }}10} \right){\rm{ }} = {\rm{ }}0\)
⇔\( x(2x + 1)(3x – 10) = 0\)
Hoặc \(x = 0\), \(x = -\frac{1}{2}\) , \(x = \frac{10}{3}\)
Vậy phương trình có 3 nghiệm.
$$ \Leftrightarrow {x_1} = - {1 \over {0,6}} = - {5 \over 3}$$