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demon311

ĐK: $x \ge \dfrac{ -1}{3}$

$\sqrt{3x+1}+\sqrt{5x+4}+x^2+2x-8=0 \\
\Leftrightarrow (\sqrt{ 3x+1}-2)+(\sqrt{ 5x+4}-3)+(x^2+2x-3)=0 \\
\Leftrightarrow \dfrac{ 3x-3}{\sqrt{ 3x+1}+2}+\dfrac{ 5x-5}{\sqrt{ 5x+4}+3}+(x-1)(x+3)=0 \\
\Leftrightarrow (x-1) \left [ \dfrac{ 3}{\sqrt{ 3x+1}+2}+\dfrac{ 5}{\sqrt{ 5x+4}+3}+(x+3) \right ]=0 \\
\Leftrightarrow x=1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $

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