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hoanghondo94

sonsac99 said:
[TEX]x^2-2(x+1)sqrt{3x+1}=2sqrt{2x^2+5x+2}-8x-5[/TEX]


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Uh , giúp được , thử thế này nhé:)

[TEX]\blue \begin{array}{l}{x^2} - 2\left( {x + 1} \right)\sqrt {3x + 1} = 2\sqrt {2{x^2} + 5x + 2} - 8x - 5 \\\Leftrightarrow {x^2} + 8x + 5 = 2\sqrt {\left( {x + 2} \right)\left( {2x + 1}\right)} + 2\left( {x + 1} \right)\sqrt {3x + 1} \\* \\2\sqrt {\left( {x + 2} \right)\left( {2x + 1} \right)} \le \left( {x + 2} \right) + \left( {2x + 1} \right) = 3x + 3 \\2\left( {x + 1} \right)\sqrt {3x + 1} \le {\left( {x + 1} \right)^2} + 3x + 1 = {x^2} + 5x + 2 \\\Rightarrow 2\sqrt {\left( {x + 2} \right)\left( {2x + 1} \right)} + 2\left( {x + 1}\right)\sqrt {3x + 1} \le {x^2} + 8x + 5 \\ \Leftrightarrow x = 1 \\\end{array}[/TEX]
 
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