pt vô tỉ

T

transformers123

$\sqrt{8x+1}=x^2+3x-1$ (ĐK: $x \ge \dfrac{-1}{8}$)
$\iff 8x+1=(x^2+3x-1)^2$
$\iff 8x+1=x^4+9x^2+1+6x^3-2x^2-6x$
$\iff x^4+6x^3+7x^2-14x=0$
$\iff x(x-1)(x^2+7x+14)=0$
$\iff \begin{cases}x=0\\x=1\\x^2+7x+14=0\ (vô\ lí)\end{cases}$
Vậy ...
 
H

huynhbachkhoa23

$\leftrightarrow \sqrt{8x+1}-3-(x^2+3x-4)=0$

$\leftrightarrow \dfrac{8(x-1)}{\sqrt{8x+1}+3}-(x-1)(x+4)=0$

$\leftrightarrow (x-1)\left( \dfrac{8}{\sqrt{8x+1}+3}-x-4 \right)=0$

$\leftrightarrow x=1$ ( Khi đặt $t=\sqrt{8x+1}$ thì $(\sqrt{8x+1}+3)(x+4)=(t+3)(t^2+4-\dfrac{1}{8}) > (t+3)(t^2+3) > 9 > 8$ )
 
H

huynhbachkhoa23

transformers123 said:
$\sqrt{8x+1}=x^2+3x-1$ (ĐK: $x \ge \dfrac{-1}{8}$)
$\iff 8x+1=(x^2+3x-1)^2$
$\iff 8x+1=x^4+9x^2+1+6x^3-2x^2-6x$
$\iff x^4+6x^3+7x^2-14x=0$
$\iff x(x-1)(x^2+7x+14)=0$
$\iff \begin{cases}x=0\\x=1\\x^2+7x+14=0\ (vô\ lí)\end{cases}$
Vậy ...
Bấm để xem đầy đủ nội dung ...

Thiếu phần này: $\sqrt{8x+1}=x^2+3x-1 \ge 0 \rightarrow x \ge \dfrac{-3+\sqrt{13}}{2}$ nên $x=0$ là bị loại.
 
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