1. $x^{3}-27\frac{x^{3}}{(x+1)^{3}}-\frac{x^{2}}{x+3}-9=0$

$2.8x^{3}+(8x^{2}-3x-6)\sqrt{x+2}=0$

3. $\sqrt{5x^{2}+14x+9}-\sqrt{x^{2}-x-20}=5\sqrt{x+1}$
-_- thật ra câu 2 rất dễ.hjc.đag định đi ngủ.k định trình bày,nghĩ lại trình bày z
đặt $\sqrt{x+2}=a =>$khi đó ptr trở thành
$8x^3+(8x^2-3a^2)a=0$
đây là đăg cấp bậc 3 rồi
giải bình thường thôi
công nhận sáng nay đệ nghĩ ra xong chửi mình ngu –  Gia Hưng 18-11-14 01:24 PM
Câu 3: ĐK:...
pt$\Leftrightarrow \sqrt{5x^2 +14x+9}=\sqrt{x^2-x-20}+5\sqrt{x+1}$
$\Leftrightarrow 5x^2+14x+9=x^2-x-20+25(x+1) +10\sqrt{(x+1)(x^2-x-20)}$
$\Leftrightarrow 5\sqrt{(x+1)(x+4)(x-5)}=2x^2-5x+2$
$\Leftrightarrow 5\sqrt{(x+4)(x^2-4x-5)}=2(x^2-4x-5)+3(x+4)$

Đặt : $\sqrt{x^2-4x-5}=a; \sqrt{x+4}=b   ; (a,b\geq 0)$
$\Rightarrow 5ab=2a^2+3b^2\Rightarrow \left[ {\begin{matrix} a = b\\ a=\frac{3}{2}b \end{matrix}} \right.$

+) $a=b\Leftrightarrow \sqrt{x^2-4x-5}=\sqrt{x+4}\Leftrightarrow x=\frac{5\pm \sqrt{61}}{2}$

+)$a=3/2b\Leftrightarrow 4x^2-25x-56=0\Leftrightarrow \left[ {\begin{matrix} x=8\\ x=-7/4 \end{matrix}} \right.$

Đối chiếu ĐK rồi kết luận
3. Bình phương 2 vế, ta có:
     $(5x^2+14x+9)-(5\sqrt{x+1}+\sqrt{x^2-x-20})^2=0$
$\Leftrightarrow 2x^2-5x+2=5\sqrt{(x-5)(x+1)(x+4)}$
$\Leftrightarrow (2x^2-5x+2)^2-25(x-5)(x+1)(x+4)=0$
$\Leftrightarrow (x-8)(4x+7)(x^2-5x-9)=0$
Với điều kiện $x \ge -1$, ta chọn được 2 nghiệm: $x\in\{8; \frac{1}{2}(5+\sqrt{61})\}$

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