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[Silverbullet069]

Giải pt: $\sqrt{\frac{x + 7}{x + 1}} + 8 = 2x^2 + \sqrt{2x - 1}$

Edited by Silverbullet069, 29-12-2016 - 12:56.

[NTA1907]

Giải pt: $\sqrt{\frac{x + 7}{x + 1}} + 8 = 2x^2 + \sqrt{2x - 1}$

ĐK: $x\geq \frac{1}{2}$

PT$\Leftrightarrow 2x^{2}-8+\left ( \sqrt{2x-1}-\sqrt{\frac{x+7}{x+1}} \right )=0$

$\Leftrightarrow 2x^{2}-8+\dfrac{2x-1-\frac{x+7}{x+1}}{\sqrt{2x-1}+\sqrt{\frac{x+7}{x+1}}}=0$

$\Leftrightarrow 2x^{2}-8+\dfrac{2x^{2}-8}{(x+1)\left ( \sqrt{2x-1}+\sqrt{\frac{x+7}{x+1}} \right )}=0$

$\Leftrightarrow (2x^{2}-8)\left ( 1+\dfrac{1}{(x+1)\left ( \sqrt{2x-1}+\sqrt{\frac{x+7}{x+1}} \right )} \right )=0$

$\Rightarrow x=2$(vì phần trong ngoặc luôn dương)