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Bài 20. Giải các phương trình:
a) \(25{x^2}-{\rm{ }}16{\rm{ }} = {\rm{ }}0\) ;
b) \(2{x^2} + {\rm{ }}3{\rm{ }} = {\rm{ }}0\)
c) \(4,2{x^2} + {\rm{ }}5,46x{\rm{ }} = {\rm{ }}0\);
d) \(4{x^2} – {\rm{ }}2\sqrt 3 x{\rm{ }} = {\rm{ }}1{\rm{ }} – {\rm{ }}\sqrt 3 \).
a) \(25{x^2}{\rm{ – }}16 = 0 \Leftrightarrow 25{x^2} = 16 \Leftrightarrow {x^2} = {\rm{ }}{{16} \over {25}}\)
\(⇔ x = ±\)\(\sqrt{\frac{16}{25}}\) = ±\(\frac{4}{5}\)
b) \(2{x^2} + {\rm{ }}3{\rm{ }} = {\rm{ }}0\). Phương trình vô nghiệm vì vế trái là \(2{x^2} + {\rm{ }}3{\rm{ }} \ge {\rm{ }}3\) còn vế phải bằng \(0\).
c) \(4,2{x^2} + {\rm{ }}5,46x{\rm{ }} = {\rm{ }}0{\rm{ }} \Leftrightarrow {\rm{ }}2x\left( {2,1x{\rm{ }} + {\rm{ }}2,73} \right){\rm{ }} = {\rm{ }}0\)
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Vậy \(x = 0\) hoặc \(2,1x{\rm{ }} + {\rm{ }}2,73{\rm{ }} = {\rm{ }}0{\rm{ }} = > {\rm{ }}x{\rm{ }} = {\rm{ }} – 1,3\).
d) \(4{x^2} – {\rm{ }}2\sqrt 3 x{\rm{ }} = {\rm{ }}1{\rm{ }} – {\rm{ }}\sqrt 3 \)
\(\Leftrightarrow {\rm{ }}4{x^2} – {\rm{ }}2\sqrt 3 x{\rm{ }}-{\rm{ }}1{\rm{ }} + {\rm{ }}\sqrt 3 {\rm{ }} = {\rm{ }}0\)
Có \(a = 4, b = -2\sqrt{3}, b’ = -\sqrt{3}, c = -1 + \sqrt{3}\)
\(\Delta’ {\rm{ }} = {\rm{ }}{\left( { – \sqrt 3 } \right)^2}-{\rm{ }}4{\rm{ }}.{\rm{ }}\left( { – 1{\rm{ }} + {\rm{ }}\sqrt 3 } \right){\rm{ }}\)
\(= {\rm{ }}3{\rm{ }} + {\rm{ }}4{\rm{ }} – {\rm{ }}4\sqrt 3 {\rm{ }} = {\rm{ }}{\left( {2{\rm{ }} – {\rm{ }}\sqrt 3 } \right)^2}\)
\({\rm{ }}\sqrt {\Delta ‘} {\rm{ }} = {\rm{ }}2{\rm{ }} – {\rm{ }}\sqrt 3 \)
\({x_1}\) = \(\frac{\sqrt{3} – 2+ \sqrt{3}}{4}\) = \(\frac{\sqrt{3} – 1}{2}\) , \({x_2}\) = \(\frac{\sqrt{3} +2 – \sqrt{3}}{4}\) = \(\frac{1}{2}\)