A<2

số cần tìm là 45

cái bím

ở lồn

bánh trưng nhé

8.

Đặt \(\left{\right. \sqrt{x^{2} + 2 x + 3} = a > 0 \\ \sqrt{x^{2} + 4 x + 5} = b > 0\) \(\Rightarrow 2 a^{2} - b^{2} = x^{2} + 1\)

Pt trở thành:

\(\sqrt{2 a^{2} - b^{2}} + 2 a = 3 b\)

\(\Leftrightarrow \sqrt{2 a^{2} - b^{2}} = 3 b - 2 a\)

\(\Rightarrow 2 a^{2} - b^{2} = 4 a^{2} - 12 a b + 9 b^{2}\)

\(\Leftrightarrow 2 a^{2} - 12 a b + 10 b^{2} = 0 \Rightarrow \left[\right. a = b \\ a = 5 b\)

\(\Rightarrow \left[\right. \sqrt{x^{2} + 2 x + 3} = \sqrt{x^{2} + 4 x + 5} \\ \sqrt{x^{2} + 2 x + 3} = 5 \sqrt{x^{2} + 4 x + 5}\)

\(\Leftrightarrow \left[\right. x^{2} + 2 x + 3 = x^{2} + 4 x + 5 \\ x^{2} + 2 x + 3 = 25 \left(\right. x^{2} + 4 x + 5 \left.\right)\)

\(\Leftrightarrow \left[\right. x = - 1 \\ 24 x^{2} + 98 x + 122 = 0 \left(\right. v n \left.\right)\)

9.

ĐKXĐ: \(- 1 \leq x \leq 1\)

Đặt \(\left{\right. \sqrt{1 + x} = a \geq 0 \\ \sqrt{1 - x} = b \geq 0\) \(\Rightarrow a^{2} + 2 b^{2} = 3 - x = - \left(\right. x - 3 \left.\right)\)

Pt trở thành:

\(a - 2 b - 3 a b = - \left(\right. a^{2} + 2 b^{2} \left.\right)\)

\(\Leftrightarrow a - 2 b + a^{2} - 3 a b + 2 b^{2} = 0\)

\(\Leftrightarrow a - 2 b + \left(\right. a - b \left.\right) \left(\right. a - 2 b \left.\right) = 0\)

\(\Leftrightarrow \left(\right. a - 2 b \left.\right) \left(\right. a - b + 1 \left.\right) = 0\)

\(\Leftrightarrow \left[\right. a = 2 b \\ a + 1 = b\)

\(\Leftrightarrow \left[\right. \sqrt{1 + x} = 2 \sqrt{1 - x} \\ \sqrt{1 + x} + 1 = \sqrt{1 - x}\)

\(\Leftrightarrow \left[\right. 1 + x = 4 \left(\right. 1 - x \left.\right) \\ x + 2 + 2 \sqrt{1 + x} = 1 - x\)

\(\Leftrightarrow \left[\right. 5 x = 3 \Rightarrow x = \frac{3}{5} \\ - 1 - 2 x = 2 \sqrt{1 + x} \left(\right. 1 \left.\right)\)

Xét (1) \(\Leftrightarrow \left{\right. - 1 - 2 x \geq 0 \\ \left(\left(\right. - 1 - 2 x \left.\right)\right)^{2} = 4 \left(\right. 1 + x \left.\right)\)

\(\Leftrightarrow \left{\right. x \leq - \frac{1}{2} \\ x^{2} = \frac{3}{4}\) \(\Rightarrow x = - \frac{\sqrt{3}}{2}\)

Vậy \(x = \left{\right. \frac{3}{5} ; - \frac{\sqrt{3}}{2} \left.\right}\)

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