1 reply to this topic

[rovklee]

Giải phương trình:
1,$\sqrt[3]{2x-1}+\sqrt[3]{x-1}=\sqrt[3]{3x+1}$
2,$3(2+\sqrt{x-2})=2x+\sqrt{x-6}$

[T M]

Giải phương trình:
1,$\sqrt[3]{2x-1}+\sqrt[3]{x-1}=\sqrt[3]{3x+1}$
2,$3(2+\sqrt{x-2})=2x+\sqrt{x-6}$

1. Cách củ chuối
$$PT\Rightarrow 2x-1+x-1+3\sqrt[3]{(2x-1)(x-1)}.\left ( \sqrt[3]{2x-1}+\sqrt[3]{x-1} \right )=3x+1 \\ \Rightarrow \sqrt[3]{(2x-1)(x-1)(3x+1)}=1 \\ \Rightarrow x=0 \vee x=\frac{7}{6}$$

Thử lại thấy cái nào thoả thì lấy !!!

------------------------------

2. Chắc là liên hợp thôi !

ĐKXĐ:

$$PT\Leftrightarrow 6-2x+3\sqrt{x-2}-\sqrt{x-6}=0\\
\Leftrightarrow -2(x-6)+3\left ( \sqrt{x-2}-2 \right )-\sqrt{x-6}=0\\
\Leftrightarrow \sqrt{x-6}\left ( -2\sqrt{x-6}+3.\frac{\sqrt{x-6}}{\sqrt{x-2}+2}-1 \right )=0\Leftrightarrow x=6$$

Vì $\left ( -2\sqrt{x-6}+3.\frac{\sqrt{x-6}}{\sqrt{x-2}+2}-1 \right )=0\Leftrightarrow 1+2\sqrt{x-6}=3.\frac{\sqrt{x-6}}{\sqrt{x-2}+2}$

Mà $\frac{3.\sqrt{x-6}}{\sqrt{x-2}+2}\leq \frac{3.\sqrt{x-6}}{2}<2\sqrt{x-6}\rightarrow \fbox{$Vo Li$}$

Edited by luxubuhl, 25-06-2012 - 20:37.