1,$4(x+1)^{2}=(2x+10)(1-\sqrt{3+2x})^{2}$
2, $5\sqrt{x}+\frac{5}{2\sqrt{x}}=2x+\frac{1}{2x}+4$
3, $x+\frac{x}{\sqrt{x^{2}-1}}=\frac{35}{12}$
1. $=> 4(x + 1)^{2} = (2x + 10)(1 + 3 + 2x - 2\sqrt{3 + 2x})$
=> $(x + 1)^{2} = (x + 5)(x + 2 - \sqrt{3 + 2x})$
=> $5x + 9 = (x + 5)\sqrt{3 + 2x}$
Đặt $a = x + 5$
$b = \sqrt{3 + 2x}$
=> $2b^{2} + a - 2 = ab$ =>...
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2. => $5(\sqrt{x} + \frac{1}{2\sqrt{x}}) = 2(x + \frac{1}{4x}) + 4$
Đặt $t = \sqrt{x} + \frac{1}{2\sqrt{x}}$
Pt => $5t = 2t^{2} + 2$ =>...
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3. => $x^{2} + \frac{x^{2}}{x^{2} - 1} + \frac{2x^{2}}{\sqrt{x}^{2} - 1} = \frac{35^{2}}{12^{2}}$
=> $\frac{x^{4}}{x^{2} - 1} + \frac{2x^{2}}{\sqrt{x^{2} - 1}} = \frac{35^{2}}{12^{2}}$
=> $(\frac{x^{2}}{\sqrt{x^{2} - 1}} + 1)^{2} =...$
=>...
- BiBi Chi yêu thích